Overview
This unit studies the gas laws that describe how gases behave under changes in pressure, volume, and temperature. Beginning with the measurable properties of gases—pressure, volume, temperature, and amount—we develop experimental and mathematical relationships such as Boyle’s law, Charles’s law, and Gay-Lussac’s law. These lead to combined gas laws and the ideal gas equation which connects all four variables and allows calculations for many practical situations. The unit also introduces concepts of kinetic molecular theory to explain macroscopic gas laws using particle ideas, as well as real-gas deviations and causes. Understanding gas laws is important because gases are involved in daily life (breathing, inflating tyres, cooking), industrial processes (gas storage, chemical reactions), and environmental phenomena. Learning these laws builds skills in laboratory measurement, graph interpretation, algebraic manipulation and applying units. The unit prepares students to solve numerical problems, design simple experiments, predict effects of changing conditions, and recognise limitations of ideal behaviour. Overall, it links observable gas behaviour to clear rules and simple molecular explanations, giving a strong foundation for later physical chemistry and thermodynamics.
Learning Objectives
- Describe and measure the basic physical properties of gases: pressure, volume, temperature and amount.
- State and apply Boyle’s law to relate pressure and volume at constant temperature.
- State and apply Charles’s law to relate volume and temperature at constant pressure.
- State and apply Gay-Lussac’s law to relate pressure and temperature at constant volume.
- Combine individual laws into the combined gas law and use it to solve problems involving changes in two or more variables.
- Use the ideal gas equation PV = nRT to calculate one variable when the others are known and to relate macroscopic measurements to amount of substance.
- Explain gas behaviour qualitatively using the kinetic molecular theory and predict effects of changing conditions.
- Recognise deviations from ideal behaviour and describe qualitatively the causes in real gases.
Topics in this chapter
17 topics · tap a topic title to jump straight to it.
Physical properties of gases: pressure, volume, temperature and amount
What are the measurable properties?
Gases are substances that fill their container and can be described by four easily measured properties. Volume is the space a gas occupies; in experiments we use litres (L) or cubic centimetres (cm3). Pressure is the force by gas on the container per unit area; common units are pascal (Pa), kilopascal (kPa), atmosphere (atm), and mm of mercury (mmHg). Temperature measures the hotness and for gas laws must be used in Kelvin (K). Amount is the quantity of gas, usually expressed in moles (mol). The behaviour of gases can be predicted by how these four properties change.
Measuring pressure and temperature
Pressure is measured by devices such as a manometer or barometer. A simple U-tube manometer compares the pressure of a gas with atmospheric pressure by the height difference of a liquid column. Thermometers measure temperature; for gas laws the scale must be converted to Kelvin by adding 273 to Celsius values.
Importance of standard conditions
To compare different experiments, we often use standard conditions such as standard temperature and pressure (STP) where T = 273 K and P = 1 atm. A mole of an ideal gas occupies 22.4 L at STP, a useful reference for calculations involving molar volume.
Units and conversions
Accuracy requires consistent units. Pressure in Pa, volume in m3, temperature in K and amount in mol fit SI. For school problems you will often see atm, L and K used with R = 0.0821 L·atm·K−1·mol−1 in the ideal gas equation. Always check and convert units before substituting into formulas.
Practical observations
Observe that when you heat a sealed balloon gently, its pressure rises slightly if volume is fixed or the balloon expands if pressure outside is fixed. Pumping air into a tyre increases its pressure and amount. These everyday examples show how changing one property affects others and lead to the gas laws developed in the following topics.
- A gas occupies 2.0 L at 1 atm; when measured, its temperature is 300 K and amount is 0.082 mol. Identify which properties are fixed and which vary.
- When a sealed syringe contains gas at room temperature and the piston is pushed in, observe that pressure increases while volume decreases.
- Measure atmospheric pressure using a simple manometer and record the result in mmHg and convert it to kPa.
- Convert 27°C to Kelvin and explain why Kelvin is necessary in gas law calculations.
- 1 atm = 101.325 kPa = 760 mmHg
- T(K) = T(°C) + 273.15
- Molar volume at STP ≈ 22.4 L·mol−1
Boyle’s Law: statement and experimental verification
Statement of Boyle’s law
Boyle’s law describes how pressure and volume of a fixed amount of gas relate at constant temperature. It states that the pressure of a given mass of gas is inversely proportional to its volume when temperature is kept constant. Mathematically we write P ∝ 1/V or PV = constant.
Historical experiment
Boyle used a J-shaped tube with mercury. A fixed amount of air trapped in the closed end changed its volume when more mercury was added. By measuring the height of mercury (which changes the pressure) and the volume of trapped air, Boyle confirmed that PV stayed nearly constant. Modern school experiments use a syringe with a pressure gauge: trapping a known amount of gas, varying the syringe volume and measuring pressure shows the inverse relation clearly.
Graphical representation
If you plot pressure P (y-axis) against 1/V (x-axis) at constant temperature you get a straight line through the origin. If you plot P against V, you get a hyperbola. Another useful plot is PV versus V; for ideal behaviour PV remains constant and so PV plotted against V is a horizontal line.
Laboratory procedure and precautions
To verify Boyle’s law: (1) trap a fixed amount of gas in a syringe fitted with a pressure gauge, (2) ensure the gas temperature is constant (perform quickly or use a water bath), (3) change the volume stepwise and record P and V pairs, (4) calculate PV for each pair. Precautions: avoid leaks, allow thermal equilibrium, and do not compress too far to avoid non-ideal effects.
Applications
Boyle’s law explains how lungs work during breathing (when chest volume increases, lung pressure falls causing air inflow) and why pressure increases in a bicycle pump when volume is reduced. It is the first of the simple gas laws and leads to the combined gas equation when temperature or amount also changes.
- A gas has volume 4.0 L at 200 kPa. What is its volume at 100 kPa at the same temperature? (Use PV = constant.)
- In a syringe the pressure reads 150 kPa when the plunger is at 20.0 cm3. If the plunger moves to 10.0 cm3, what is the new pressure?
- A trapped air bubble in a deep well decreases in volume as pressure increases. Explain using Boyle’s law.
- Measurements: given P-V pairs (100 kPa, 2.0 L), (200 kPa, 1.0 L) show PV is constant.
- Boyle's law: P ∝ 1/V (at constant T, n)
- PV = constant (for fixed n and T)
Charles’s Law: volume–temperature relation
Statement of Charles’s law
Charles’s law deals with how the volume of a fixed amount of gas changes with temperature at constant pressure. It states that volume is directly proportional to absolute temperature (in Kelvin). In symbols: V ∝ T or V/T = constant (at constant pressure and amount). This means if the absolute temperature is doubled, the volume doubles under the same pressure.
Experimental verification and method
A typical school experiment uses a syringe whose tip is open to the atmosphere so pressure remains constant. Measure the gas volume at several temperatures: for example in an ice bath (0°C), room temperature (about 25°C) and a warm water bath (about 60°C). Record the volumes after thermal equilibrium is reached. When plotted, the volumes increase linearly with temperature in °C; plotting against Kelvin gives a straight line passing through the origin, showing V ∝ T and confirming the proportionality to absolute temperature.
Importance of absolute temperature and absolute zero
When the measured volume is extrapolated backward to zero volume on a V versus T(°C) graph, the intercept appears near −273°C. This empirical result supports the existence of absolute zero where molecular motion would cease. Therefore Charles’s law requires temperature in Kelvin for mathematical accuracy; use T(K) = T(°C) + 273.15 when applying formulas.
Graphical forms and slopes
Plotting V (y-axis) against T (K) (x-axis) results in a straight line whose slope equals V/T = constant for the sample. If different masses of gas are used the slope changes because volume at a given temperature depends on amount. The straight-line nature provides a simple way to evaluate data quality and estimate experimental error by checking linearity and intercepts.
Applications and examples in everyday life
Charles’s law explains why hot-air balloons expand when heated: the heated air inside increases in volume at the same pressure, producing buoyancy. It also explains why tyres or inflatable toys expand slightly when warmed and shrink in cold weather. In cooking, air pockets in batters expand on heating due to increased volume of trapped gases, affecting texture.
Limitations and practical cautions
Ensure pressure is constant during the experiment; an open system to atmosphere or a piston that can move freely is required. Large temperature ranges may introduce non-ideal gas behaviour or thermal expansion of apparatus, so keep ranges moderate and allow proper equilibrium. Use Kelvin in calculations to avoid errors and always label axes with units when plotting.
- A balloon has volume 3.0 L at 300 K. What is its volume at 360 K if pressure remains constant?
- Air in a sealed container at 20°C occupies 2.5 L. Find its volume at 80°C at constant pressure (convert to Kelvin).
- Explain why Charles’s law requires temperature in Kelvin rather than Celsius.
- Plotting measured V versus T(°C) for a gas gives a straight line; extrapolate to find temperature at which V would be zero.
- Charles's law: V ∝ T (at constant P, n)
- V/T = constant (for fixed n and P)
Gay-Lussac’s Law: pressure–temperature relation
Statement and definition
Gay-Lussac’s law (also known as Amontons' law in some texts) concerns the relationship between pressure and temperature for a fixed amount of gas held at constant volume. It states that the pressure of a gas is directly proportional to its absolute temperature (measured in Kelvin). In equation form: P ∝ T or P/T = constant when volume and amount are fixed.
Experimental setup and verification
To test Gay-Lussac’s law in the laboratory use a rigid vessel fitted with a pressure gauge so volume cannot change. Place the vessel in water baths at different temperatures and read pressure after equilibrium. Typical points might be at 0°C, 25°C and 50°C. Convert temperatures to Kelvin and plot P versus T (K). If the points fall on a straight line through the origin, the proportionality is verified and the constant P/T can be calculated.
Molecular explanation
From the kinetic molecular point of view, temperature reflects the average kinetic energy of gas particles. When temperature rises, particles move faster and collide with the container walls with greater momentum and more frequently. At fixed volume this increases the force per unit area on the walls, raising measured pressure in direct proportion to temperature. This microscopic picture helps link measurable quantities to particle behaviour.
Safety and practical considerations
Heating a sealed container increases pressure and can be dangerous if the container is not rated for the pressures reached. Use metal or purpose-built vessels with pressure relief where possible, and do not heat glass containers without safety measures. Take readings gradually, and avoid rapid temperature changes that can cause pressure spikes or instrument damage.
Calculations and rearrangements
For two states of the same gas at constant volume: P1/T1 = P2/T2. Use this relation to predict pressure changes when temperature changes. Rearranged forms allow solving for any unknown: P2 = P1 × T2/T1. Always convert temperatures to Kelvin before substituting. Check results for reasonableness; pressure should fall if temperature falls, and vice versa.
Real-world examples and significance
This law explains why aerosol cans or pressurised containers should not be left in hot places: heating increases internal pressure and may cause rupture. It also explains the pressure increase in car engines when combustion raises temperature in cylinders at near-constant volume. In weather balloons and industrial systems, understanding pressure–temperature links is essential for design and safety.
- A fixed volume gas has pressure 150 kPa at 300 K. What will be its pressure at 360 K?
- Explain why heating a sealed can may cause it to burst in terms of Gay-Lussac’s law.
- Record pressure readings of a gas in a rigid flask at two temperatures and show P/T is constant.
- An aerosol can warns against exposure to temperatures above 50°C; explain the reason using pressure–temperature relation.
- Gay-Lussac's law: P ∝ T (at constant V, n)
- P/T = constant (for fixed n and V)
Combined Gas Law and practical problem solving
Derivation and statement
The combined gas law merges Boyle’s, Charles’s and Gay-Lussac’s laws to handle simultaneous changes in pressure, volume and temperature for a fixed amount of gas. Combining PV = constant at constant T, V/T = constant at constant P, and P/T = constant at constant V gives the relation PV/T = constant for a given mass of gas. Thus for an initial state (P1, V1, T1) and final state (P2, V2, T2) we write P1V1/T1 = P2V2/T2.
When to use it
Use the combined law when more than one property changes and the amount of gas remains constant. It is especially useful in laboratory or real-life situations where temperature, pressure and volume change together — for example inflating a tyre which warms up, or calculating the change in pressure in a sealed container when volume and temperature change.
Problem-solving steps
1. Convert all temperatures to Kelvin. 2. Ensure pressure and volume units are consistent (same units both sides). 3. Identify which quantity is unknown. 4. Apply P1V1/T1 = P2V2/T2 and solve algebraically. 5. Check units and whether the result is physically reasonable (e.g., positive volume, pressure).
Worked numerical tips
Rearrange to find the unknown: P2 = P1V1T2/(T1V2), V2 = P1V1T2/(P2T1), T2 = P2V2T1/(P1V1). Beware that if the gas amount changes (leak or addition), you must use the ideal gas equation instead. Also remember that small compressions or expansions that change temperature need measurement or estimation of final temperature before applying the combined law.
Significance
The combined law unifies the three simple laws and forms a bridge to the ideal gas equation PV = nRT. It allows students to handle real multi-step situations and strengthens skills in algebra, unit management and physical interpretation. Many examination questions require applying this law to real situations and to check limiting cases, such as when one variable is fixed and the equation reduces to a simple law.
- A gas at 1.0 atm and 2.0 L at 300 K is compressed to 1.5 L and heated to 360 K. Find the final pressure using P1V1/T1 = P2V2/T2.
- Air in a bicycle pump: initial P = 100 kPa, V = 0.5 L at 293 K; after compression V = 0.1 L, temperature rises to 320 K. Calculate final pressure.
- A balloon at 20°C (1.0 atm, 5.0 L) is taken to a higher altitude where pressure is 0.8 atm and temperature is 250 K. Find new volume.
- Combined gas law: P1V1/T1 = P2V2/T2
- Rearranged forms: P2 = P1V1T2/(T1V2), V2 = P1V1T2/(P2T1), T2 = P2V2T1/(P1V1)
The Ideal Gas Equation: PV = nRT
Statement and meaning
The ideal gas equation PV = nRT relates pressure P, volume V, amount n (in moles), temperature T (in Kelvin) and the universal gas constant R. It provides a single relation that combines the simple gas laws and is widely used to calculate unknown variables when the gas behaves ideally. The equation assumes an ideal gas: particles have negligible volume and no intermolecular forces. Under ordinary conditions many gases approximate this behaviour closely.
Choice and value of R
The value of R depends on the units chosen for pressure and volume. Common classroom choices are R = 0.08206 L·atm·K−1·mol−1 when using litres and atmospheres, and R = 8.314 J·K−1·mol−1 when using SI units (Pa and m3) because 1 J = 1 Pa·m3. Decide the units before calculations and keep units consistent: if P is in atm use the 0.08206 value, if P is in Pa use R = 8.314 and V in cubic metres.
Algebraic rearrangements and uses
From PV = nRT you can rearrange to find any unknown: n = PV/(RT), V = nRT/P and T = PV/(nR). Use these forms to convert measured volume to moles or to predict volume changes when temperature or pressure changes. Always convert temperature to Kelvin before substituting in the formula. The equation is central in quantitative gas problems and connects laboratory measurements with amounts of substance used in stoichiometry.
Molar volume and STP
Applying the ideal gas equation for n = 1 at standard temperature and pressure (273.15 K and 1 atm) yields the molar volume: V = RT/P = 0.08206 × 273.15 / 1 ≈ 22.414 L. This number is useful when converting volumes at STP to moles quickly without full calculation. If textbooks or teachers use 1 bar instead of 1 atm, molar volume changes slightly; be careful which standard is used.
Accuracy and limitations
The ideal gas equation is an approximation. At high pressures and low temperatures interactions and finite molecular size lead to errors. For accurate engineering calculations introduce correction factors (e.g., van der Waals equation). However, for most school problems at moderate pressure and temperature the ideal equation gives results accurate enough and helps develop problem-solving skills linking chemistry and mathematics.
Practical tips
Check units and convert where necessary: volumes in litres for R = 0.08206 or m3 for R = 8.314, pressures in atm or Pa accordingly, and temperature in Kelvin. Dimensional checking after calculation helps spot mistakes: PV should have units matching nRT. Present answers with correct significant figures and units to make your solution clear in examinations.
- Calculate the number of moles in 11.2 L of a gas at STP using PV = nRT with R = 0.08206 L·atm·K−1·mol−1.
- A 5.0 L container holds nitrogen gas at 2.0 atm and 300 K. Find the number of moles using n = PV/(RT).
- Find the volume occupied by 0.50 mol of oxygen at 1.0 atm and 298 K using V = nRT/P.
- Ideal gas equation: PV = nRT
- n = PV/(RT), V = nRT/P, T = PV/(nR)
Molar volume and STP calculations
Molar volume concept
Molar volume is the volume occupied by one mole of a gas under specified conditions. At standard temperature and pressure (STP: 273.15 K and 1 atm) the molar volume of an ideal gas is about 22.414 litres per mole. This value follows directly from PV = nRT when n = 1.
Using molar volume in problems
When you are given a volume of gas at STP, you can easily find the number of moles by dividing the volume by 22.414 L·mol−1. Conversely, the mass of the gas can be found by multiplying moles by molar mass. For example, to find mass of oxygen in a 44.8 L sample at STP: moles = 44.8 / 22.414 = 2.0 mol; mass = 2.0 × 32.00 g = 64.0 g.
Standard conditions variations
Be aware that different texts sometimes use different standard conditions (e.g., STP defined at 273 K and 1 bar rather than 1 atm). If using 1 bar (100 kPa) the molar volume is slightly different (≈ 22.711 L·mol−1). Always use the value consistent with the pressure unit used in PV = nRT.
Practical examples
Molar volume is useful in gas collection experiments where volume of gas produced is measured at known temperature and pressure and converted to moles. In chemical reactions involving gases it simplifies stoichiometric calculations at STP: you can equate volumes directly if all gases are measured under the same conditions because equal moles occupy equal volumes.
Unit practice
Remember to use correct units: if you use R = 0.08206 L·atm·K−1·mol−1 then P must be in atm and V in litres. Convert temperatures to Kelvin. If pressures are given in kPa convert to atm (1 atm = 101.325 kPa) or use R in kPa·L·K−1·mol−1 = 8.314 kPa·L·K−1·mol−1/1000? (better to use R = 8.314 J·K−1·mol−1 with SI units P in Pa and V in m3). Careful unit management prevents arithmetic errors.
- Find the number of moles in 44.8 L of gas at STP and then find its mass if the gas is CO2 (M = 44.01 g·mol−1).
- A reaction releases 11.2 L of hydrogen gas at STP. How many moles and what mass of H2 is produced?
- If molar volume at 273 K and 1 bar is 22.711 L, find moles in 2.27 L at these conditions.
- Molar volume at STP (1 atm, 273.15 K): Vm ≈ 22.414 L·mol−1
- n = V / Vm (when gas at STP)
Kinetic molecular theory: molecular explanation of gas laws
Basic postulates
The kinetic molecular theory (KMT) explains gas behaviour using simple ideas about particles. Key postulates: (1) a gas consists of a large number of tiny particles (atoms or molecules) in constant random motion; (2) the particles are separated by distances much larger than their own size, so the volume of particles is negligible; (3) collisions between particles and with container walls are perfectly elastic (no net energy loss); (4) there are no attractive or repulsive forces between particles except during collisions; (5) the average kinetic energy of particles is directly proportional to the absolute temperature.
Connecting KMT to gas laws
Boyle’s law: reducing volume makes particles collide more often with the walls, increasing pressure. Charles’s law: increasing temperature raises average kinetic energy, so particles strike walls with more momentum and the volume must increase at constant pressure. Gay-Lussac’s law: at fixed volume hotter particles exert more pressure due to greater momentum transfer on collisions. The KMT gives a microscopic basis for the proportionalities seen in macroscopic gas laws.
Mean kinetic energy and temperature
Temperature is a measure of average translational kinetic energy of particles. Although we do not compute kinetic energy in class 9, the idea helps interpret why absolute zero (0 K) corresponds to vanishing particle motion and why gases condense when cooled: particles slow down and inter-particle forces become significant.
Limitations and simplifications
KMT assumes point particles and no interactions, so it cannot explain real gas deviations at high pressures or low temperatures where particle size and intermolecular forces matter. However, for common conditions it provides a clear physical picture linking particle motion to pressure, volume and temperature and helps students visualise why gas laws work.
Classroom models and demonstrations
Demonstrations—such as using ping-pong balls in a box to show random motion, or a molecular model of elastic collisions—help visualise KMT. Reinforce that KMT is a model: useful for explanation but built on simplifying assumptions that will be refined in later studies.
- Use particle diagrams to explain why pressure increases when volume decreases at constant temperature.
- Explain qualitatively why heating a gas increases its pressure in a sealed container using particle motion ideas.
- Describe what happens to particle speed and collision frequency when temperature is lowered.
Real gases and deviations from ideal behaviour
Overview of deviations
Real gases do not always follow the ideal gas equation exactly because the ideal model simplifies molecular behaviour. Two main causes of deviation are the finite volume of molecules and intermolecular forces. At high pressures molecules are crowded so their own volume reduces the free volume for motion. At low temperatures or when molecules are polar, attractive forces become significant and can reduce pressure below that predicted by the ideal equation or lead to condensation into liquids.
How to see deviations experimentally
One way to detect non-ideal behaviour is to calculate the compressibility factor Z = PV/(nRT). For an ideal gas Z = 1. Measured Z values that differ from unity indicate non-ideal effects: Z > 1 typically at high pressures due to repulsion and finite size, Z < 1 in a region where attractions dominate. Plotting Z versus pressure at a given temperature gives a curve that departs from the horizontal line Z = 1 and shows where ideal assumptions break down.
Physical interpretation of deviations
At low pressures and high temperatures, molecular distances are large and kinetic energy dominates, so the ideal approximation works well. As pressure rises, molecules are closer and excluded volume becomes important; they cannot be treated as point particles. At lower temperatures particle speed is reduced and attractive forces pull molecules together, reducing momentum transfer to container walls and thus lowering pressure compared with the ideal prediction. In extreme cases condensation occurs and the gas law no longer applies at all for the two-phase system.
Corrections and van der Waals idea
To correct for real behaviour, more sophisticated equations such as the van der Waals equation introduce parameters for molecular size (b) and attraction (a): (P + a(n/V)2)(V − nb) = nRT. While this is beyond Class 9 detail, the idea shows how simple corrections can improve predictions by accounting for finite size and attractions. These parameters are specific to each gas and determined experimentally.
Practical significance
Deviations matter in industrial and laboratory settings: calculations for high-pressure gas cylinders, liquefaction processes, or low-temperature storage require non-ideal corrections. For common classroom conditions (near room temperature and moderate pressure) small molecules such as N2 and O2 obey the ideal gas law closely, so ideal approximations are appropriate for most problems in Class 9.
Identifying when to worry
Always consider the pressure and temperature ranges. If pressure is much greater than 1 atm or temperature is near condensation points, expect non-ideal behaviour. Discuss qualitatively why results differ from ideal predictions and link observations to molecular causes; this reasoning is valued in exams and strengthens understanding of the limits of models.
- Explain why high-pressure storage of CO2 requires corrections to ideal gas calculations.
- A gas shows PV/(nRT) = 0.95 at certain conditions; discuss what this implies about intermolecular forces.
- Describe why oxygen behaves nearly ideally at room temperature but not near its liquefaction point.
Units, dimensional analysis and using R correctly
Why units matter
Gas law calculations involve quantities with units: pressure, volume, temperature and amount. The gas constant R links these quantities numerically, but its numerical value depends on the units chosen. Using inconsistent units (for example pressure in kPa while using R in L·atm·K−1·mol−1) will give incorrect numerical results. Therefore a clear plan of units before calculation is essential.
Common choices of R and their units
Two values of R are most commonly used in school problems. R = 0.08206 L·atm·K−1·mol−1 is convenient when volumes are in litres and pressure in atmospheres. R = 8.314 J·K−1·mol−1 is used with SI units where pressure is in pascals (Pa) and volume in cubic metres (m3), because in SI, 1 Pa·m3 = 1 J. Some teachers prefer R = 8.314 kPa·L·K−1·mol−1/1000 conversions; stick with the common two values and ensure consistency.
Dimensional analysis to check work
Dimensional analysis means checking that both sides of an equation have the same units. For PV = nRT in SI units, P (Pa) × V (m3) has units of Pa·m3 which equals J; nRT uses R in J·K−1·mol−1 × K × mol giving J. This confirms unit consistency. If using R = 0.08206, PV will be in L·atm; nRT will be numerically equal when P is in atm, V in L, T in K and n in mol. Dimensional checks help spot unit mistakes before calculating.
Common unit conversions
Be comfortable converting between pressure units: 1 atm = 101.325 kPa = 101325 Pa = 760 mmHg. Convert volume between litres and cubic metres: 1 L = 1×10−3 m3. Convert temperatures by T(K) = T(°C) + 273.15. Treat unit conversion as a standard step in problem solving so that values plugged into formulas are consistent with the chosen R.
Practical examples and pitfalls
If a problem gives pressure in kPa and volume in L, either convert pressure to atm to use R = 0.08206 or convert volume to m3 and pressure to Pa to use R = 8.314. A common pitfall is forgetting to convert temperature to Kelvin: using Celsius with PV = nRT breaks the proportionality and produces incorrect answers. Another error is mismatching R and units; always write R with its units on the answer sheet to reduce mistakes.
Exam and lab tips
In exams show unit conversions explicitly when space allows. In lab reports include units for measured values and the R value used. When the teacher or question statement indicates a specific R, use it; otherwise choose the R that makes unit handling easiest for the given data. Clear unit management demonstrates good scientific practice and earns marks in assessments.
- Convert 2.0 atm to kPa and Pa, and convert 5.0 L to cubic metres.
- Show that PV and nRT have the same units when using SI units and R = 8.314 J·K−1·mol−1.
- Given pressure in mmHg and volume in litres, convert pressure to atm for use with R = 0.08206 L·atm·K−1·mol−1.
- 1 atm = 101.325 kPa = 101325 Pa = 760 mmHg
- 1 L = 10−3 m3
- T(K) = T(°C) + 273.15
Gas mixtures and partial pressures (Dalton’s Law)
Statement of Dalton’s law
Dalton’s law of partial pressures states that in a mixture of non-reacting gases the total pressure equals the sum of the partial pressures of the individual gases. Each gas behaves as if it alone occupied the container at the same temperature, so the pressure contribution of each gas adds up to the observed total pressure.
Derivation from ideal gas idea
Using the ideal gas equation for each component: PiV = niRT. For a mixture the total pressure Ptotal satisfies PV = ntotal RT where ntotal = n1 + n2 + ... . Therefore Ptotal = (n1 + n2 + ...)RT/V = P1 + P2 + ... . This follows directly from PV = nRT and the assumption that gases do not chemically react or strongly interact, so their pressures are additive.
Partial pressure and mole fraction
The partial pressure of a component is proportional to its mole fraction in the mixture: Pi = xi Ptotal where xi = ni/ntotal. This relation is useful because knowing composition by amount tells us how much each gas contributes to the total pressure. For example, air at 1 atm has oxygen partial pressure about 0.21 atm because oxygen is about 21% by mole in dry air.
Collecting gas over water: vapour pressure correction
A common laboratory situation is collecting a gas over water. The measured pressure in the collection vessel equals the sum of the dry gas pressure and the vapour pressure of water at that temperature. To find the pressure of the dry collected gas subtract the known vapour pressure of water: Pgas = Ptotal − PH2O. Use vapour pressure tables for the experiment temperature to make this correction accurately.
Applications and physiological relevance
Dalton’s law is important in respiratory physiology: oxygen and other gases exert partial pressures that drive diffusion into blood. It also helps in calculating the composition of industrial gas mixtures, predicting behaviour in gas blending, and estimating hazards where a toxic gas may reach a dangerous partial pressure even if its overall fraction is small.
Limitations and practical notes
Dalton’s law assumes ideal behaviour and negligible interactions between different gases. At very high pressures or with reactive or condensable gases, interactions may affect partial pressures. For most classroom and atmospheric situations the law is accurate and provides a straightforward tool for solving mixture problems.
- A 2.0 L container holds 1.0 mol N2 and 0.5 mol O2 at 300 K. Find total pressure using PV = nRT.
- Air at 1.00 atm contains about 21% O2 by mole. What is the partial pressure of O2?
- A gas collected over water shows total pressure 100.0 kPa at 25°C. Vapour pressure of water is 3.17 kPa. Find the dry gas pressure.
- Dalton's law: Ptotal = ΣPi
- Pi = xi Ptotal where xi = ni/ntotal
- PiV = niRT for each gas
Applications: tyres, syringes, balloons and breathing
Tyres: pressure, temperature and safety
Tyres are an everyday example where gas laws are useful. When you pump air into a tyre you increase the amount n, leading to higher pressure for the tyre's fixed volume. Driving warms tyres and increases temperature; by Gay-Lussac’s law pressure rises if volume is nearly constant. Understanding these links explains why recommended tyre pressures are checked when tyres are cold and why significant temperature rises can risk overpressure and blowouts.
Syringes and Boyle’s law in practice
A syringe with nozzle closed demonstrates Boyle’s law: reducing the plunger volume increases pressure inside, and releasing it lowers pressure. This principle has medical and laboratory uses, for example when drawing fluids or creating small pressure differences. In experiments, a syringe with a pressure gauge is a simple tool to verify P ∝ 1/V at constant temperature.
Balloons and Charles’s law
Hot-air balloons rise because heating the air inside increases its volume at constant external pressure, reducing density and creating buoyant force. Simple balloons expand when heated and contract when cooled, which is Charles’s law in action. These effects are also relevant in packaging and storage where gas-filled objects change with environmental temperature.
Breathing, partial pressures and physiology
Breathing relies on pressure and partial pressure differences. Inhalation increases thoracic volume, lowering lung pressure below atmospheric and causing air inflow (Boyle’s law). Oxygen uptake in lungs depends on partial pressure differences: oxygen moves from alveolar air (higher PO2) to blood (lower PO2). At high altitude lower total pressure reduces oxygen partial pressure and can cause breathlessness. Dalton’s law helps calculate these partial pressures and their physiological effects.
Problem solving with combined laws
Real-world problems often require combining laws: tyre pressure changes need PV = nRT if both temperature and amount change; balloon behaviour uses V ∝ T; syringe and lung examples use Boyle’s law. Identify which variables change and which remain nearly constant to choose the correct relation. Also account for safety: sealed systems heated rapidly can overpressurise; always consider material limits and pressure relief mechanisms.
Engineering and everyday relevance
From designing pressure vessels to understanding home heating effects on tyre pressure, gas laws are practical. They provide quantitative tools to predict behaviour and guide safe usage. Encourage students to connect classroom calculations to these real applications to appreciate the usefulness of gas law concepts.
- A car tyre has pressure 220 kPa at 20°C. After driving it warms to 40°C. Assuming tyre volume and amount constant, find the new pressure.
- Explain why a balloon shrinks overnight when temperature drops using Charles’s law.
- Inhale-exhale demonstration: explain pressure–volume changes in lungs during quiet breathing using Boyle’s law and atmospheric pressure.
- PV = nRT to relate pressure, volume, temperature and amount in application problems
- Use P1V1/T1 = P2V2/T2 for multi-variable changes
Experimental methods: measuring gas laws in the lab
Designing experiments
To study gas laws you need apparatus to measure pressure, volume and temperature while keeping one or more variables constant. Typical equipment includes syringes with pressure gauges, gas syringes with stopcocks, U-tube manometers, thermostatted water baths for temperature control, rigid flasks with pressure sensors, and data-loggers for precise readings. Choose apparatus to match the law under test: varied volume at constant temperature for Boyle’s law, varied temperature at constant pressure for Charles’s law, and so on.
Common procedures
Boyle’s law: trap a fixed amount of gas in a syringe at room temperature; record P and V while compressing the syringe in steps; ensure thermal equilibrium. Charles’s law: use a gas in a graduated bore or a syringe open to atmosphere to keep pressure constant; place in water baths at different temperatures and record V. Gay-Lussac: use a rigid vessel with a pressure gauge; heat to different temperatures and record P.
Controls and accuracy
Control variables: ensure the amount of gas is fixed (no leaks), temperature is stable for Boyle’s law experiments, and pressure is constant for Charles’s law tests. Repeat measurements and average to reduce random error. Use appropriate units and convert temperature to Kelvin for analysis. Account for systematic errors such as instrument calibration offsets and thermal expansion of apparatus if precise data are required.
Data presentation
Present data in tables with columns for P, V, T and calculated quantities like PV. Plot relevant graphs: P vs 1/V for Boyle’s law, V vs T (K) for Charles’s law. Straight lines or constant values indicate agreement with the laws. Discuss deviations and possible experimental causes such as leaks, heat exchange or inaccurate readings.
Safety and practical tips
When heating sealed containers use safety shields and rated apparatus. Avoid over-pressuring systems. Use gloves and eye protection for hot baths. Dispose of chemicals safely and follow laboratory rules. Good record-keeping and careful measurement make experiments reproducible and help in learning how gas laws apply in practice.
- Outline a step-by-step experimental plan to verify Boyle’s law using a syringe and pressure gauge.
- Describe how to measure Charles’s law using a gas syringe in water baths of known temperatures.
- List precautions to ensure accurate gas law measurements and how to spot a leak in apparatus.
Graphical analysis of gas law data
Choosing the right plot
Graphing helps verify gas laws and extract constants. For Boyle’s law plot P on the y-axis and 1/V on the x-axis; a straight line passing through the origin indicates P ∝ 1/V. For Charles’s law plot V on the y-axis against T in Kelvin on the x-axis to get a straight line through origin. For Gay-Lussac’s law plot P versus T (K). For the ideal gas equation, plotting PV against T for fixed n gives a straight line with slope nR.
Extracting constants
From plots you can determine constants like the proportionality constant in Boyle’s law (PV) or the gas constant R from slope measurements. For example, if you plot PV versus T for a fixed n, the slope equals nR, so dividing slope by n gives R. Ensure units are consistent so the numerical value of R matches expected values.
Interpreting scatter and deviations
Experimental points rarely lie exactly on theory lines. Use best-fit lines and discuss sources of scatter: measurement uncertainty, leaks, non-ideal behaviour, thermal lag, parallax errors in reading volumes. If systematic deviation is observed (e.g., PV changes with V), discuss possible physical reasons such as non-ideal gas effects or instrument bias.
Graphical extrapolation
Graphical extrapolation is used in Charles’s law to find absolute zero by extending the V vs T(°C) line to the temperature where volume would become zero. While this is an ideal extrapolation, it shows the conceptual link to absolute zero and supports using Kelvin scale for calculations.
Presentation and lab report tips
Label axes with variable and units, include error bars if possible, state the best-fit equation and the correlation coefficient to show fit quality. Discuss whether experimental constant values (e.g., PV) match theoretical expectations within error margins. Clear graphical analysis strengthens conclusions in laboratory reports and examinations.
- Given experimental data pairs for P and V, plot P vs 1/V and determine whether the data support Boyle’s law.
- From a V vs T (°C) plot extrapolate to find approximate absolute zero and compare with −273°C.
- Given PV vs T data for a fixed amount of gas, determine R from the slope.
Stoichiometry with gases: reacting volumes and gas law use
Volume–mole relationship in gases
When gases react under the same temperature and pressure, their volumes are directly proportional to the numbers of moles reacting. This is because equal numbers of moles occupy equal volumes under identical conditions (Avogadro’s principle). Thus simple volume ratios can represent stoichiometric ratios in reactions involving only gases measured at the same conditions.
Using PV = nRT for differing conditions
If gases in a reaction are measured under different temperatures or pressures, convert volumes to moles using PV = nRT before applying stoichiometry. The formula n = PV/(RT) converts a measured gas volume to the amount in moles so you can combine it with mole ratios from chemical equations. This step is essential when conditions are not identical or when the gas is collected at non‑standard conditions.
Worked approach for gas reactions
Typical steps: (1) write the balanced chemical equation and note mole ratios; (2) convert measured gas volumes to moles using PV = nRT (with correct R and units) or to STP equivalents using molar volume; (3) use mole ratios to find amounts of other reactants or products; (4) convert moles back to volumes or masses as required. Pay attention to whether gases are at STP so you can use the 22.4 L per mole shortcut safely.
Examples and common classroom problems
For the reaction 2H2 + O2 → 2H2O (gas), 2 volumes of hydrogen react with 1 volume oxygen at the same conditions. If 22.4 L H2 at STP are consumed, 11.2 L O2 at STP are required. For reactions producing gases where the gas is collected over water, subtract water vapour pressure to find the dry gas pressure before converting to moles. Also account for limiting reagents: the reactant that provides fewer moles according to stoichiometry limits the amount of product.
Practical tips and pitfalls
Remember that volume ratios only apply when gases are measured at the same temperature and pressure. A common mistake is equating volumes without checking conditions. Another pitfall is unit mismatch when using PV = nRT—always convert temperatures to Kelvin and ensure R matches units. In exam answers show the conversions and the steps clearly to earn full credit.
Importance in chemistry
Gas stoichiometry links laboratory measurements to theoretical mole calculations and is used in industrial gas production, laboratory synthesis, and environmental calculations (e.g., CO2 emissions). Mastering these methods helps solve a wide range of chemical problems involving gases and prepares students for more advanced stoichiometry and thermochemistry topics.
- Hydrogen and oxygen react as 2H2 + O2 → 2H2O. If 22.4 L of H2 at STP is used, what volume of O2 at STP is required?
- A reaction produces 11.2 L of CO2 at 300 K and 1 atm. Find moles of CO2 and the mass produced.
- Given measured gas volumes at different temperatures, convert to moles using PV = nRT before applying stoichiometry.
- n = PV/(RT) to convert volume to moles
- At same P and T: V1/V2 = n1/n2 (volumes proportional to moles)
Temperature scales and absolute zero
Common temperature scales
The two temperature scales most used in chemical calculations are Celsius (°C) and Kelvin (K). Celsius is convenient for everyday use because it references the freezing and boiling points of water (0°C and 100°C). Kelvin is the absolute scale used in gas laws and thermodynamics because its zero point corresponds to absolute zero, the theoretical temperature where particle motion in the classical picture would stop.
Conversion and why Kelvin is required
Convert Celsius to Kelvin using T(K) = T(°C) + 273.15. Kelvin must be used in gas law formulas because these laws relate quantities to absolute temperature: for example V ∝ T and P ∝ T are valid only when T is absolute. Using Celsius would make proportionalities meaningless because Celsius zero is arbitrary and not physically linked to molecular energy.
Estimating absolute zero experimentally
One classroom demonstration uses Charles’s law data plotted as V versus T(°C). Extrapolating the straight line to the temperature where volume would be zero gives a value close to −273°C. Although the experiment is approximate due to measurement error and non-ideal gas behaviour, it historically contributed to recognising an absolute temperature scale and provides a useful conceptual link between measured volumes and particle motion.
Physical meaning of absolute zero
Absolute zero (0 K or −273.15°C) is the lower limit of the thermodynamic temperature scale. In classical terms it represents a state where molecular translational motion ceases; in modern physics quantum effects remain but energy reaches its lowest permitted value. For gas laws, absolute zero is the point where extrapolated volume or pressure would become zero, reinforcing why Kelvin is the correct scale for proportional relations.
Practical tips and common conversions
Always convert temperatures to Kelvin before substituting into PV = nRT or combined gas law expressions. For temperature differences, Celsius and Kelvin scales have the same step size, so a change of 1°C equals 1 K in difference. When answering exam questions show the conversion explicitly to avoid loss of marks. Remember that negative Celsius values are common but Kelvin values must be positive for physically meaningful results in gas calculations.
Broader relevance
Understanding absolute zero prepares students for later topics in thermodynamics and physical chemistry where temperature links to internal energy and entropy. It also helps in interpreting behaviour of gases near very low temperatures where quantum effects and condensation become important.
- Convert 25°C and −40°C to Kelvin and explain why Kelvin is used in gas law equations.
- Explain how an experiment plotting V versus T(°C) can be used to estimate absolute zero.
- If volume doubles when temperature rises from 300 K to 600 K at constant pressure, what is the initial and final temperature in °C?
- T(K) = T(°C) + 273.15
Safety, environmental and industrial relevance of gases
Safety with gases
Many gases present hazards: some are flammable (hydrogen, methane), others toxic (carbon monoxide), and some are simple asphyxiants (nitrogen) that displace breathable air. High-pressure gas cylinders also pose mechanical risks. Knowledge of gas laws helps predict how pressure changes with temperature so users avoid overheating cylinders or sealed containers. Correct storage, securing cylinders upright, using appropriate regulators and checking for leaks are basic safety practices.
Handling and emergency precautions
Never expose pressurised containers to heat sources since Gay-Lussac’s law predicts pressure rises with temperature, which can lead to rupture. When checking for leaks use soapy water to see bubbles rather than a flame. Ensure good ventilation for gases heavier than air (like CO2) since they can accumulate at floor level causing asphyxiation. Keep safety data sheets handy and use personal protective equipment as required.
Environmental significance
Gases play key roles in climate and air quality. Greenhouse gases such as carbon dioxide and methane trap heat in the atmosphere; understanding partial pressures and concentrations is important in monitoring and modelling climate change. Pollutant gases at certain partial pressures cause health hazards and environmental damage; gas laws help convert measured concentrations into partial pressures and assess exposure risks.
Industrial applications
Industries use gas laws in many ways: designing reactors and pressure vessels uses PV = nRT and corrections for non-ideal behaviour; refrigeration and compressed natural gas systems depend on pressure–temperature relations; and processes like gas liquefaction require understanding deviations from ideality. Engineers apply advanced equations and safety factors to ensure reliable operation under real conditions.
Storage and transportation
Compressed gas storage must account for temperature changes during transport. For example, a cylinder heated in sun may experience pressure increase; regulators and pressure relief valves mitigate risk. Liquefied gases require cryogenic storage and insulation because expansion on warming is hazardous; liquid nitrogen and liquid oxygen systems are designed with venting to avoid overpressure.
Societal and ethical considerations
Using gases responsibly includes reducing greenhouse emissions, managing industrial releases, and ensuring safe consumer products (like aerosol cans). Knowledge of gas behaviour helps make informed choices and supports policy discussions on environmental protection and public safety. Teaching gas laws with real-world safety and environmental context prepares students to act responsibly in science and daily life.
- Explain why leaving an aerosol can in sunlight is dangerous using Gay-Lussac’s law.
- Describe how CO2 fire extinguishers work and why understanding gas pressure is important for their safe use.
- Discuss why storage of liquid nitrogen requires special insulated containers due to very low temperatures and gas expansion on warming.
Key Concepts
- Pressure
- Force exerted by gas per unit area on the walls of its container.
- Volume
- Space occupied by a gas, commonly measured in litres or cubic metres.
- Temperature (Kelvin)
- Measure of average kinetic energy of gas particles; Kelvin scale starts at absolute zero.
- Amount (mole)
- Quantity of substance measured in moles; one mole contains Avogadro’s number of particles.
- Boyle’s law
- At constant temperature, pressure of a fixed gas is inversely proportional to its volume (PV = constant).
- Charles’s law
- At constant pressure, volume of a fixed gas is directly proportional to its absolute temperature (V/T = constant).
- Gay-Lussac’s law
- At constant volume, pressure of a fixed gas is directly proportional to its absolute temperature (P/T = constant).
- Combined gas law
- Relation P1V1/T1 = P2V2/T2 for a fixed amount of gas undergoing a change of state.
- Ideal gas equation
- PV = nRT, relating pressure, volume, amount and temperature for an ideal gas.
- Molar volume
- Volume occupied by one mole of gas under specified conditions; about 22.414 L·mol−1 at STP.
- Dalton’s law
- Total pressure of a gas mixture equals the sum of the partial pressures of its components.
- Kinetic molecular theory
- Model that explains gas laws by assuming particles in random motion with elastic collisions and negligible volume.
- Absolute zero
- Theoretical temperature (0 K or −273.15°C) at which particles have minimal classical motion.
- Non-ideal gas
- A real gas that deviates from ideal behaviour due to finite particle volume and intermolecular forces.
- Universal gas constant (R)
- Proportionality constant in PV = nRT; value depends on chosen units (e.g., 0.08206 L·atm·K−1·mol−1).
- Partial pressure
- Pressure a component gas would exert if it alone occupied the whole container at the same temperature.
- STP
- Standard temperature and pressure commonly taken as 273.15 K and 1 atm for gas calculations.
Practice Questions
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A 2.00 L sample of gas at 1.50 atm is compressed to 1.20 L at constant temperature. What is the final pressure? / 1.50 atm पर 2.00 L गैस को स्थिर तापमान पर 1.20 L तक संपीड़ित किया जाता है। अंतिम दाब क्या होगा?
Show answer
Using Boyle's law P1V1 = P2V2. P2 = P1V1/V2 = 1.50 atm × 2.00 L / 1.20 L = 2.50 atm. / बॉयल के नियम के अनुसार P1V1 = P2V2. P2 = 1.50 atm × 2.00 L / 1.20 L = 2.50 atm।
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A balloon has volume 4.0 L at 27°C. What will be its volume at 127°C if pressure is constant? / 27°C पर 4.0 L वॉलून का आयतन 127°C पर क्या होगा यदि दाब स्थिर है?
Show answer
Convert to Kelvin: T1 = 300 K, T2 = 400 K. Use Charles's law V1/T1 = V2/T2 so V2 = V1 × T2/T1 = 4.0 L × 400/300 = 5.33 L. / केल्विन में बदलें: T1 = 300 K, T2 = 400 K. V2 = 4.0 × 400/300 = 5.33 L।
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A rigid container holds gas at 100 kPa and 300 K. If the temperature is raised to 450 K, what is the new pressure? / एक कठोर कंटेनर में गैस 100 kPa और 300 K पर है। यदि तापमान 450 K कर दिया जाए, तो नया दाब क्या होगा?
Show answer
Use P/T = constant (Gay-Lussac). P2 = P1 × T2/T1 = 100 kPa × 450/300 = 150 kPa. / Gay-Lussac के अनुसार P2 = 100 kPa × 450/300 = 150 kPa।
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One mole of an ideal gas at 298 K occupies what volume at 1.00 atm? Use R = 0.08206 L·atm·K−1·mol−1. / 1.00 atm पर 298 K में 1 मोल आदर्श गैस का आयतन कितना होगा? R = 0.08206 L·atm·K−1·mol−1 मानें।
Show answer
Use V = nRT/P. V = 1×0.08206×298/1.00 = 24.45 L (approximately). / V = nRT/P = 1×0.08206×298 = 24.45 L (लगभग)।
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Air contains about 21% oxygen by volume. What is the partial pressure of oxygen in air at total pressure 760 mmHg? / वायु में लगभग 21% ऑक्सीजन आयतन के अनुसार होती है। कुल दाब 760 mmHg पर ऑक्सीजन का आंशिक दाब क्या होगा?
Show answer
Partial pressure PO2 = 0.21 × 760 mmHg = 159.6 mmHg ≈ 160 mmHg. / PO2 = 0.21 × 760 mmHg = 159.6 mmHg ≈ 160 mmHg।
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A reaction produces 11.2 L of hydrogen at STP. How many moles of hydrogen are produced? / एक अभिक्रिया STP पर 11.2 L हाइड्रोजन उत्पन्न करती है। कितने मोल हाइड्रोजन उत्पन्न होते हैं?
Show answer
At STP 1 mol gas = 22.4 L. Moles = 11.2/22.4 = 0.50 mol. / STP पर 1 मोल गैस = 22.4 L. moles = 11.2/22.4 = 0.50 mol।
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A mixture contains 2.0 mol N2 and 1.0 mol O2 in a 10.0 L container at 300 K. Find the total pressure. / एक मिश्रण में 10.0 L कंटेनर में 2.0 mol N2 और 1.0 mol O2 300 K पर हैं। कुल दाब ज्ञात कीजिए।
Show answer
Total n = 3.0 mol. Use PV = nRT with R = 0.08206 L·atm·K−1·mol−1: P = nRT/V = 3.0×0.08206×300 / 10.0 = 7.3854 /10 = 0.7385 atm ≈ 0.739 atm (or ×101.325 = 74.9 kPa). / कुल n = 3.0 mol. P = 3.0×0.08206×300 /10.0 = 0.7385 atm ≈ 0.739 atm (≈74.9 kPa)।
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A syringe with trapped air reads 50.0 mL at 101 kPa. When the plunger is pressed the volume becomes 20.0 mL. What is the new pressure (assume temperature constant)? / 101 kPa पर 50.0 mL गैस दिखाने वाली सिरिंज में प्लंजर दबाने पर आयतन 20.0 mL हो जाता है। नया दाब क्या होगा? तापमान स्थिर मानें।
Show answer
By Boyle's law P1V1 = P2V2. P2 = P1V1/V2 = 101 kPa × 50.0/20.0 = 252.5 kPa. / P2 = 101 × 50/20 = 252.5 kPa।
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Why must temperature be in Kelvin when using gas law formulas? / गैस नियमों का उपयोग करते समय तापमान केल्विन में क्यों होना चाहिए?
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Because the gas laws relate properties to absolute temperature which is proportional to average kinetic energy; Kelvin scale has zero at absolute zero making proportionalities (e.g., V ∝ T) valid. Celsius has an arbitrary zero and would give incorrect results. / गैस नियम वास्तविक तापमान (औसत गतिज ऊर्जा के अनुपात) से सम्बंधित हैं; केल्विन शून्य पर सापेक्ष शून्य रखता है इसलिए V ∝ T आदि सही रहते हैं। सेल्सियस का शून्य यादृच्छिक है और त्रुटिपूर्ण परिणाम देगा।
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A sealed cylinder of gas at 20°C has pressure 2.00 atm. If the temperature becomes −40°C, what is the new pressure? / 20°C पर एक सीलबंद सिलेंडर में गैस का दाब 2.00 atm है। यदि तापमान −40°C हो जाए तो नया दाब क्या होगा?
Show answer
Convert to Kelvin: T1 = 293.15 K, T2 = 233.15 K. Use P2 = P1 × T2/T1 = 2.00 × 233.15/293.15 = 2.00 × 0.795 = 1.59 atm (approx). / केल्विन में: T1 = 293.15 K, T2 = 233.15 K. P2 = 2.00 × 233.15/293.15 ≈ 1.59 atm।
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