Gas Laws

ChemistryStates Of MatterFor JEE aspirants

Gas laws are the simple rules that link the pressure ( $p$ ), volume ( $V$ ), temperature ( $T$ ) and amount ( $n$ ) of a gas. Boyle, Charles, Gay-Lussac and Avogadro each hold two variables fixed; together they give the ideal gas equation $p V = n R T$ . Dalton's law handles mixtures and Graham's law handles diffusion. These gas laws work because gas molecules barely attract one another. The topic is part of the JEE Advanced syllabus (States of Matter: Gases and Liquids).

On this page1Gas variables2Boyle3Charles4Gay-Lussac5Avogadro6pV = nRT7Density and M8Dalton9Graham10Examples

Key Formulas - Quick Reference

★ Must learn Boyle ( $T$ , $n$ fixed): $p_{1} V_{1} = p_{2} V_{2}$ . Charles ( $p$ , $n$ fixed): $\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$ . Gay-Lussac ( $V$ , $n$ fixed): $\frac{p _{1}}{T _{1}} = \frac{p _{2}}{T _{2}}$ .
★ Must learn Ideal gas equation: $p V = n R T$ , with $R = 8.314 J K^{- 1} mol^{- 1}$ , or $0.08314 L bar K^{- 1} mol^{- 1}$ , or $0.0821 L atm K^{- 1} mol^{- 1}$ .
Combined gas law (fixed $n$ ): $\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}$ ; if $n$ changes, use $\frac{p _{1} V _{1}}{n _{1} T _{1}} = \frac{p _{2} V _{2}}{n _{2} T _{2}}$ .
★ Must learn Density and molar mass: $d = \frac{pM}{R T}$ , so $M = \frac{d R T}{p}$ ; also $M = 2 \times$ vapour density.
★ Must learn Dalton: $p_{total} = p_{1} + p_{2} + \dots$ and $p_{i} = x_{i} p_{total}$ ; moist gas: $p_{dry} = p_{total} - aqueous tension$ .
★ Must learn Graham: $\frac{r _{1}}{r _{2}} = \frac{M _{2}}{M _{1}} = \frac{d _{2}}{d _{1}}$ (same $p$ , $T$ ); in general $r \propto \frac{p}{M}$ .
Molar volume: $22.7$ L at STP (273.15 K, 1 bar); $22.4$ L at 273.15 K, 1 atm; $24.8$ L at SATP (298.15 K, 1 bar).
Temperature: $T / K = t /^{\circ} C + 273.15$ ; absolute zero $= 0 K = - 273.15^{\circ} C$ .

1. The Gaseous State and Its Variables

In a gas the molecules are far apart and the forces between them are very weak. That is why every gas, whatever its chemistry, follows the same few laws. At ordinary conditions only eleven elements are gases: $H_{2}$ , $N_{2}$ , $O_{2}$ , $F_{2}$ , $C l_{2}$ and the six noble gases (He, Ne, Ar, Kr, Xe, Rn).

Highly compressible: most of a gas is empty space.
No fixed shape or volume: a gas fills its whole container.
Uniform pressure: a gas pushes equally on every wall.
Low density compared with liquids and solids.
Complete mixing: gases mix in all proportions without stirring.

Four measurable quantities fix the state of a gas: pressure, volume, temperature and amount. Every gas law is a relation between them.

Figure 1: A gas is fully described by

p

V

T

and

n

. The atmosphere holds up a 760 mm column of mercury, so

1 atm = 760 mm Hg = 101.325 kPa = 1.01325 bar

. Temperature in every gas law is in kelvin.

2. Boyle's Law (Pressure and Volume)

Boyle's law (1662): at constant temperature, the pressure of a fixed amount of gas is inversely proportional to its volume.

p \propto \frac{1}{V} (T, n fixed) \Rightarrow p V = k_{1}, p_{1} V_{1} = p_{2} V_{2}

The constant $k_{1}$ equals $n R T$ , so it depends on the amount of gas, its temperature and the units used. Halve the pressure and the volume doubles. A plot of $p$ against $V$ at one temperature is a rectangular hyperbola called an isotherm; a hotter isotherm lies higher. A plot of $p$ against $1/ V$ is a straight line through the origin.

Figure 2: Boyle's law for

1

mol of gas, drawn from

p = \frac{n R T}{V}

. At

V = 10

L the pressures are

1.66

3.33

and

4.99

bar at 200, 400 and 600 K. Each curve is an isotherm; the higher the curve, the higher the temperature.

Real data show the same thing. For $0.09$ mol of $C O_{2}$ at 300 K the product $p V$ stays close to $n R T = 224.5$ Pa m $^{3}$ over a five-fold change of pressure:

$p$ / $1 0^{4}$ Pa	$V$ / $1 0^{- 3}$ m $^{3}$	$p V$ / Pa m $^{3}$
2.0	11.20	224.0
2.5	8.92	223.0
3.5	6.42	224.7
4.0	5.63	225.2
6.0	3.74	224.4
8.0	2.81	224.8
10.0	2.24	224.0

2.1 Density and pressure

Squeezing a gas packs the same molecules into less space, so the gas gets denser. With $d = m / V$ and $V = k_{1} / p$ :

d = (\frac{m}{k _{1}}) p = k^{'} p

At constant temperature the density of a gas is directly proportional to its pressure. Examiners often test Boyle's law through such less obvious plots:

Figure 3: Graph questions test the same law in disguise.

p V = n R T

is constant, so

p V

p

is flat;

lo g p = lo g (n R T) - lo g V

has slope

- 1

; and

d = \frac{pM}{R T}

is a line through the origin that is steeper at lower

T

At high pressures real gases stop obeying Boyle's law, and the

p

1/ V

plot curves. The reasons are on the Deviation From Ideal Gas Behaviour page.

Key idea

Boyle: $p V$ is constant at fixed

T

and

n

. Every Boyle graph is a hyperbola, a flat line, a slope of

- 1

or a line through the origin.

3. Charles's Law and the Kelvin Scale

Charles and Gay-Lussac, working on hot-air balloons, found that at constant pressure a gas expands by $\frac{1}{273.15}$ of its volume at $0^{\circ} C$ for every degree rise in temperature:

V_{t} = V_{0} (1 + \frac{t}{273.15}) = V_{0} (\frac{273.15 + t}{273.15})

Defining a new scale $T = 273.15 + t$ (the Kelvin, absolute or thermodynamic scale, written without a degree sign) turns this into a direct proportion:

Charles's law: at constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature.

V = k_{2} T

, or

\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}

A $V$ vs $t$ plot at fixed pressure is a straight line called an isobar. Isobars at different pressures have different slopes, but extended backwards they all reach $V = 0$ at $- 273.15^{\circ} C$ . This lowest possible temperature is absolute zero. No gas reaches it: every real gas liquefies first. Charles's law is best obeyed at low pressure and high temperature.

Figure 4: Isobars for

1

mol of gas. At

0^{\circ} C

the volumes are

22.7

11.4

and

5.7

L at 1, 2 and 4 bar; each line rises by

\frac{1}{273.15}

of that value per degree. Lower pressure gives a steeper line, and every line points to absolute zero,

- 273.15^{\circ} C

Exam Trick Doubling the Celsius temperature does not double the volume. Heating from

27^{\circ} C

54^{\circ} C

raises

V

only by the factor

\frac{327}{300} = 1.09

. Convert to kelvin before taking any ratio.

Key idea

Charles: $V / T$ is constant at fixed

p

and

n

, with

T

in kelvin. All isobars point to absolute zero.

4. Gay-Lussac's Law (Pressure and Temperature)

Tyre pressure rises on a hot afternoon and falls on a cold morning, while the tyre volume hardly changes. Gay-Lussac expressed this as:

Gay-Lussac's law: at constant volume, the pressure of a fixed amount of gas is directly proportional to its absolute temperature.

\frac{p}{T} = k_{3}

, or

\frac{p _{1}}{T _{1}} = \frac{p _{2}}{T _{2}}

It follows directly from Boyle's and Charles's laws. A $p$ vs $T$ line at fixed volume is an isochore; the smaller the volume, the steeper the isochore.

Figure 5: Isochores for

1

mol of gas:

p = \frac{n R}{V} T

. At 300 K the pressures are

2.49

1.25

and

0.62

bar for 10, 20 and 40 L. The lines pass through the origin only when

T

is in kelvin; against

^{\circ}

C they cut the axis at

- 273.15^{\circ} C

like the isobars.

Law	Held constant	Relation	Graph name
Boyle	$T$ , $n$	$p V$ = constant	isotherm
Charles	$p$ , $n$	$V / T$ = constant	isobar
Gay-Lussac	$V$ , $n$	$p / T$ = constant	isochore
Avogadro	$p$ , $T$	$V / n$ = constant	(straight line $V$ vs $n$ )

5. Avogadro's Law (Volume and Amount)

Avogadro's law (1811): equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

V \propto n

, or

V = k_{4} n

One mole contains $N_{A} = 6.022 \times 1 0^{23}$ molecules, so one mole of any ideal gas occupies the same volume at a given temperature and pressure. At STP (273.15 K and 1 bar, the current IUPAC standard) this molar volume is $22.71$ L mol $^{- 1}$ .

Figure 6: Equal volumes, equal numbers of molecules. Molecule size and mass do not matter because the molecules fill almost none of the space. Molar volume:

22.7

L at STP (1 bar),

22.4

L at 1 atm and

0^{\circ} C

24.8

L at SATP (298.15 K, 1 bar).

Gas	Ar	$C O_{2}$	$N_{2}$	$O_{2}$	$H_{2}$	Ideal gas
Molar volume at STP / L	22.37	22.54	22.69	22.69	22.72	22.71

Real gases come within 1.5% of the ideal value. Because $n = m / M$ , Avogadro's law also gives $M = k_{4} \frac{m}{V} = k_{4} d$ : at fixed temperature and pressure, the density of a gas is proportional to its molar mass.

Three standards appear in problems. Old STP: 273.15 K and 1 atm, molar volume

22.414

L. IUPAC STP: 273.15 K and 1 bar,

22.711

L. SATP: 298.15 K and 1 bar,

24.789

L. Use whichever the question defines; if it says only "STP" in an older-style problem, 22.4 L is usually intended.

Quick Recall: tap to check

A gas is heated from

27^{\circ} C

327^{\circ} C

at constant pressure. Factor by which

V

changes?

\frac{600}{300} = 2

: the volume doubles.

Which law gives an isochore?

Gay-Lussac's law:

p \propto T

at constant

V

Why does 1 mol of

H_{2}

and 1 mol of

C O_{2}

occupy the same volume at STP?

Equal numbers of molecules and negligible molecular volume: the volume depends only on

n

T

and

p

, not on the kind of molecule.

6. The Ideal Gas Equation

Combining the three proportionalities gives one equation that holds for any ideal gas:

Figure 7: Three proportionalities, one equation. Match

R

to the units of

p

and

V

: with bar and L use

0.08314

, with atm and L use

0.0821

, with Pa and

m^{3}

(or energy) use

8.314

V \propto \frac{n T}{p} \Rightarrow p V = n R T

$R$ is the same for every gas, so it is called the universal gas constant. Its value follows from the molar volume at STP:

R = \frac{p V}{n T} = \frac{( 1 0 ^{5} Pa ) ( 22.71 \times 1 0 ^{- 3} m ^{3} )}{( 1 mol ) ( 273.15 K )} = 8.314 J K^{- 1} mol^{- 1}

An ideal gas is a hypothetical gas that obeys Boyle's, Charles's and Avogadro's laws exactly. Its molecules have no attractions and no volume of their own. Real gases come close to it only at low pressure and high temperature. Because $p V = n R T$ fixes the state of a gas from four variables, it is also called the equation of state.

6.1 Combined gas law

For a fixed amount of gas moving from state 1 to state 2, $\frac{p V}{T} = n R$ is the same in both states:

\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}

Any one of the six quantities can be found from the other five. Boyle's, Charles's and Gay-Lussac's laws are just this equation with one pair cancelled.

6.2 Density and molar mass of a gas

Put $n = m / M$ into $p V = n R T$ and divide by $V$ :

\frac{m}{V} = d = \frac{pM}{R T} \Rightarrow M = \frac{d R T}{p}

Measuring the density of a gas at known $p$ and $T$ gives its molar mass. Vapour density is the density of a gas relative to hydrogen at the same $T$ and $p$ , so $VD = \frac{M}{2}$ and $M = 2 \times VD$ .

Exam Trick Never memorise six versions of the gas laws. Write $\frac{p _{1} V _{1}}{n _{1} T _{1}} = \frac{p _{2} V _{2}}{n _{2} T _{2}}$ and strike out whatever the question keeps constant. It also handles leaks, refills and gas driven out by heating, where

n

changes.

JEE Advanced Connected vessels. When two flasks are joined by a tube, the pressure becomes the same everywhere, but each flask may sit at its own temperature. The total number of moles is conserved:

\frac{p _{1} V _{1}}{R T _{1}} + \frac{p _{2} V _{2}}{R T _{2}} = \frac{p V _{1}}{R T _{1}^{'}} + \frac{p V _{2}}{R T _{2}^{'}}

Example: two equal bulbs hold a gas at 1 bar and 300 K. One bulb is heated to 600 K while the other stays at 300 K. Then

\frac{2}{300} = p (\frac{1}{300} + \frac{1}{600})

, so

p = 1.33

bar, and the hot bulb holds only one-third of the gas.

Key idea

p V = n R T

is the master equation: every other gas law, and

d = pM / R T

, is a rearrangement of it.

7. Dalton's Law of Partial Pressures

Dalton's law (1801): the total pressure of a mixture of non-reacting gases equals the sum of the partial pressures of the individual gases. The partial pressure of a gas is the pressure it would exert if it alone filled the same volume at the same temperature.

p_{total} = p_{1} + p_{2} + p_{3} + \dots (T, V fixed)

Each gas obeys $p_{i} = \frac{n _{i} R T}{V}$ , and all share the same $R T / V$ . Dividing one by the total gives the mole fraction $x_{i} = \frac{n _{i}}{n}$ :

\frac{p _{i}}{p _{total}} = \frac{n _{i}}{n _{1} + n _{2} + n _{3}} = x_{i} \Rightarrow p_{i} = x_{i} p_{total}

Figure 8: Each non-reacting gas pushes on the walls as if it were alone, so

p = p_{A} + p_{B}

. The count of molecules decides the share:

p_{i} = x_{i} p

. (Numbers from NCERT Exercise 5.8: 0.5 L of

H_{2}

at 0.8 bar and 2.0 L of

O_{2}

at 0.7 bar moved into 1 L.)

7.1 Gas collected over water

Gases prepared in the lab are often collected by displacing water, so they are saturated with water vapour. The pressure of that saturated vapour is the aqueous tension. It depends only on temperature, and it must be subtracted:

p_{dry gas} = p_{total} - aqueous tension

Figure 9: A gas collected over water is moist. With the levels equal,

p_{dry gas} = p_{atm} - aqueous tension

. At 298 K the correction is about

0.031

bar (

23.8

mm Hg), about 3% of the total.

Exam Trick In one container, pressure ratio = mole ratio, never mass ratio. Convert grams to moles first: 1 g of

H_{2}

and 16 g of

O_{2}

give

p_{H_{2}} : p_{O_{2}} = 0.5 : 0.5 = 1 : 1

, not

1 : 16

Key idea

Dalton: gases in a mixture act independently; each partial pressure is its mole fraction times the total.

8. Graham's Law of Diffusion

Gases spread into one another (diffusion) and escape through tiny holes (effusion). In 1831 Thomas Graham found that light gases do both faster.

Diffusion

Mixing of one gas into another because of random molecular motion, for example the smell of a perfume spreading through a room. Slowed by collisions with the other gas.

Effusion

Escape of gas molecules through a pinhole into a vacuum, one molecule at a time, for example a balloon slowly going flat. Graham's law holds exactly for effusion.

Graham's law: at the same temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (or of its density).

\frac{r _{1}}{r _{2}} = \frac{M _{2}}{M _{1}} = \frac{d _{2}}{d _{1}} and in general \frac{r _{1}}{r _{2}} = \frac{p _{1}}{p _{2}} \frac{M _{2}}{M _{1}}

Rate is the volume (or moles) that passes per unit time. For the same volume, the time taken is proportional to $M$ . The classic demonstration uses ammonia and hydrogen chloride:

Figure 10: Lighter gas, faster diffusion.

\frac{r _{N H_{3}}}{r _{HCl}} = \frac{36.5}{17} = 1.47

, so in a 100 cm tube the ring forms

100 \times \frac{1.47}{2.47} = 59.4

cm from the ammonia end, nearer the HCl end.

Why Graham's law holds. Molecules escape through a hole at a rate proportional to their average speed, and the average speed at a given temperature is

\frac{8 R T}{π M} \propto \frac{1}{M}

(see Kinetic Molecular Theory of Gases). The law is used to enrich uranium:

^{235} U F_{6}

effuses faster than

^{238} U F_{6}

by a factor of only

352/349 = 1.0043

, so the gas is passed through more than a thousand stages.

A handy check: $H_{2}$ effuses four times faster than $O_{2}$ , because $32/2 = 4$ . If a question gives times, flip the ratio: $\frac{t _{1}}{t _{2}} = \frac{M _{1}}{M _{2}}$ for equal volumes.

Quick Recall: tap to check

0.2 mol

O_{2}

and 0.8 mol

N_{2}

are at a total pressure of 5 bar.

p_{O_{2}}

x_{O_{2}} = 0.2

, so

p_{O_{2}} = 0.2 \times 5 = 1

bar.

A gas diffuses

\frac{1}{2}

as fast as

O_{2}

. Its molar mass?

\frac{1}{2} = \frac{32}{M}

, so

M = 128

g mol

^{- 1}

Hydrogen collected over water at 298 K reads 1.000 bar. Pressure of dry

H_{2}

Aqueous tension at 298 K is about

0.031

bar, so

p_{H_{2}} \approx 0.969

bar.

9. Choosing the Right Law

Almost every gas-law numerical fits one of four patterns. Decide the pattern first, then convert units.

Figure 11: Three questions pick the law. Most mistakes come before this chart: temperature left in

^{\circ}

C, or

R

in units that do not match

p

and

V

9.1 The whole concept at a glance

Figure 12: The whole concept on one page. Every branch is a special case of

p V = n R T

except Graham's law, which comes from molecular speeds.

10. Solved Examples

Solved Example 1

A balloon filled with hydrogen at room temperature bursts if the pressure falls below 0.2 bar. At 1 bar the gas occupies 2.27 L. Up to what volume can the balloon expand?

Solution:

Temperature and amount are fixed, so Boyle's law applies: $p_{1} V_{1} = p_{2} V_{2}$ .

V_{2} = \frac{p _{1} V _{1}}{p _{2}} = \frac{1 bar \times 2.27 L}{0.2 bar} = 11.35 L

The balloon must stay below 11.35 L.

Solved Example 2

On a ship in the Pacific Ocean at

23.4^{\circ} C

, a balloon is filled with 2 L of air. What is its volume when the ship reaches the Indian Ocean at

26.1^{\circ} C

Solution:

Pressure and amount are fixed, so $\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$ with $T_{1} = 296.4$ K and $T_{2} = 299.1$ K.

V_{2} = \frac{V _{1} T _{2}}{T _{1}} = \frac{2 L \times 299.1 K}{296.4 K} = 2.018 L

Solved Example 3

A gas occupies 600 mL at

25^{\circ} C

and 760 mm Hg. What is its pressure at a height where the temperature is

10^{\circ} C

and the volume is 640 mL?

Solution:

Combined gas law with $T_{1} = 298$ K, $T_{2} = 283$ K:

p_{2} = \frac{p _{1} V _{1} T _{2}}{T _{1} V _{2}} = \frac{760 \times 600 \times 283}{298 \times 640} mm Hg = 676.6 mm Hg

Solved Example 4

A neon-dioxygen mixture contains 70.6 g

O_{2}

and 167.5 g Ne. The total pressure is 25 bar. Find the partial pressure of each gas (

M_{Ne} = 20

g mol

^{- 1}

Solution:

$n_{O_{2}} = \frac{70.6}{32} = 2.206$ mol and $n_{Ne} = \frac{167.5}{20} = 8.375$ mol, total $10.581$ mol.

$x_{O_{2}} = \frac{2.206}{10.581} = 0.2085$ and $x_{Ne} = 1 - 0.2085 = 0.7915$ .

$p_{O_{2}} = 0.2085 \times 25 = 5.21$ bar; $p_{Ne} = 0.7915 \times 25 = 19.79$ bar. (Rounding $x$ to 0.21 and 0.79 first gives 5.25 and 19.75 bar.)

Solved Example 5

A flask of air at

27^{\circ} C

is heated on a flame by mistake until it reaches

477^{\circ} C

. What fraction of the air is driven out? (NCERT Exercise 5.11)

Solution:

The flask is open, so $p$ and $V$ stay fixed and $n T$ is constant: $\frac{n _{2}}{n _{1}} = \frac{T _{1}}{T _{2}} = \frac{300}{750} = 0.4$ .

Fraction expelled $= 1 - 0.4 = 0.6$ , that is 60% of the air.

Solved Example 6

Pay load is the mass of air displaced minus the mass of the balloon (including its gas). Find the pay load of a balloon of radius 10 m and mass 100 kg filled with helium at 1.66 bar and

27^{\circ} C

. Density of air

= 1.2

kg m

^{- 3}

R = 0.083

bar dm

^{3}

^{- 1}

mol

^{- 1}

. (NCERT Exercise 5.16)

Solution:

$V = \frac{4}{3} π r^{3} = \frac{4}{3} π (10)^{3} = 4188.8$ m $^{3} = 4.1888 \times 1 0^{6}$ dm $^{3}$ .

Mass of air displaced $= 1.2 \times 4188.8 = 5026.5$ kg.

n_{He} = \frac{p V}{R T} = \frac{1.66 \times 4.1888 \times 1 0 ^{6}}{0.083 \times 300} = 2.793 \times 1 0^{5} mol

$m_{He} = 4 \times 2.793 \times 1 0^{5}$ g $= 1117$ kg.

Pay load $= 5026.5 - (100 + 1117) = 3809$ kg, about $3.81 \times 1 0^{3}$ kg.

Solved Example 7

A mixture of

H_{2}

and

O_{2}

at 1 bar contains 20%

H_{2}

by mass. Find the partial pressure of

H_{2}

. (NCERT Exercise 5.19)

Solution:

Take 100 g: $n_{H_{2}} = \frac{20}{2} = 10$ mol, $n_{O_{2}} = \frac{80}{32} = 2.5$ mol.

$x_{H_{2}} = \frac{10}{12.5} = 0.8$ , so $p_{H_{2}} = 0.8 \times 1 = 0.8$ bar. Only 20% of the mass, but 80% of the pressure.

Solved Example 8

1 g of an ideal gas A at

27^{\circ} C

exerts 2 bar in a flask. When 2 g of another ideal gas B is added at the same temperature, the pressure becomes 3 bar. Relate their molar masses. (NCERT Exercise 5.5)

Solution:

Same $V$ and $T$ , so $p_{i} = \frac{m _{i}}{M _{i}} \frac{R T}{V}$ . Here $p_{A} = 2$ bar and $p_{B} = 3 - 2 = 1$ bar.

\frac{p _{A}}{p _{B}} = \frac{m _{A} / M _{A}}{m _{B} / M _{B}} \Rightarrow 2 = \frac{1/ M _{A}}{2/ M _{B}} = \frac{M _{B}}{2 M _{A}} \Rightarrow M_{B} = 4 M_{A}

Solved Example 9

Under the same conditions, 50 mL of

O_{2}

effuses through a pinhole in 20 s. How long will 50 mL of

C H_{4}

take?
(A) 10 s
(B) 14.1 s
(C) 28.3 s
(D) 40 s

Solution:

Answer: (B). For equal volumes, $\frac{t _{C H_{4}}}{t _{O_{2}}} = \frac{16}{32} = 0.707$ , so $t = 0.707 \times 20 = 14.1$ s. The lighter gas is faster, so the time must be less than 20 s.

Solved Example 10

For a fixed amount of an ideal gas, which plot is NOT a straight line?
(A)

p

1/ V

at constant

T

(B)

V

t

(

^{\circ}

C) at constant

p

(C)

p

V

at constant

T

(D)

lo g p

lo g V

at constant

T

Solution:

Answer: (C). $p V$ = constant makes $p$ vs $V$ a rectangular hyperbola. (A) is a line through the origin, (B) is a line that cuts the axis at $- 273.15^{\circ} C$ , and (D) is a line of slope $- 1$ .

Solved Example 11

The density of a gas at 2 bar and

0^{\circ} C

equals that of

N_{2}

at 5 bar and

0^{\circ} C

. Find its molar mass. (NCERT Exercise 5.4)

Solution:

$d = \frac{pM}{R T}$ with the same $d$ and $T$ : $p_{1} M_{1} = p_{2} M_{2}$ .

$2 M = 5 \times 28$ , so $M = 70$ g mol $^{- 1}$ .

Practice Questions

Minimum pressure to compress 500 dm $^{3}$ of air at 1 bar to 200 dm $^{3}$ at $30^{\circ} C$ ? (Ex. 5.1)Answer: $p_{2} = \frac{1 \times 500}{200} = 2.5$ bar.
A 120 mL vessel holds gas at $35^{\circ} C$ and 1.2 bar. It is moved into a 180 mL vessel at $35^{\circ} C$ . New pressure? (Ex. 5.2)Answer: $p_{2} = \frac{1.2 \times 120}{180} = 0.8$ bar.
Using $p V = n R T$ , show that at a given temperature the density of a gas is proportional to its pressure. (Ex. 5.3)Answer: $d = \frac{m}{V} = \frac{pM}{R T}$ ; with $M$ , $R$ , $T$ fixed, $d \propto p$ .
Volume of $H_{2}$ at $20^{\circ} C$ and 1 bar released when 0.15 g Al reacts with caustic soda (Drainex). (Ex. 5.6)Answer: $2Al + 2NaOH + 6 H_{2} O \to$ $2Na [Al (OH)_{4}] + 3 H_{2}$ ; 0.00833 mol $H_{2}$ , $V = 0.203$ L = 203 mL.
Pressure of 3.2 g $C H_{4}$ + 4.4 g $C O_{2}$ in a 9 dm $^{3}$ flask at $27^{\circ} C$ ? (Ex. 5.7)Answer: $n = 0.2 + 0.1 = 0.3$ mol; $p = 0.831$ bar $= 8.31 \times 1 0^{4}$ Pa.
0.5 L of $H_{2}$ at 0.8 bar and 2.0 L of $O_{2}$ at 0.7 bar go into a 1 L vessel at $27^{\circ} C$ . Total pressure? (Ex. 5.8)Answer: $p_{H_{2}} = 0.4$ bar, $p_{O_{2}} = 1.4$ bar, total 1.8 bar.
Density of a gas is 5.46 g dm $^{- 3}$ at $27^{\circ} C$ and 2 bar. Density at STP? (Ex. 5.9)Answer: $d_{2} = 5.46 \times \frac{1}{2} \times \frac{300}{273} = 3.00$ g dm $^{- 3}$ .
34.05 mL of phosphorus vapour weighs 0.0625 g at $546^{\circ} C$ and 0.1 bar. Molar mass? (Ex. 5.10)Answer: $M = \frac{m R T}{p V} = 1.25 \times 1 0^{3}$ g mol $^{- 1}$ with the printed data. At 1 bar it would be 125 g mol $^{- 1}$ , which matches $P_{4}$ (124).
Temperature of 4.0 mol of gas in 5 dm $^{3}$ at 3.32 bar ( $R = 0.083$ )? (Ex. 5.12)Answer: $T = \frac{3.32 \times 5}{4 \times 0.083} = 50$ K.
Total pressure of 8 g $O_{2}$ + 4 g $H_{2}$ in 1 dm $^{3}$ at $27^{\circ} C$ ( $R = 0.083$ )? (Ex. 5.15)Answer: $n = 0.25 + 2 = 2.25$ mol; $p = 56.0$ bar.
Volume of 8.8 g $C O_{2}$ at $31.1^{\circ} C$ and 1 bar ( $R = 0.083$ )? (Ex. 5.17)Answer: $n = 0.2$ mol; $V = 0.2 \times 0.083 \times 304.1 = 5.05$ L.
2.9 g of a gas at $95^{\circ} C$ occupies the same volume as 0.184 g $H_{2}$ at $17^{\circ} C$ at the same pressure. Molar mass? (Ex. 5.18)Answer: $\frac{2.9}{M} \times 368 = \frac{0.184}{2} \times 290$ , so $M = 40$ g mol $^{- 1}$ .
SI unit of $p V^{2} T^{2} / n$ ? (Ex. 5.20)Answer: Pa m $^{6}$ K $^{2}$ mol $^{- 1}$ $=$ N m $^{4}$ K $^{2}$ mol $^{- 1}$ .
Two bulbs of 2 L (at 1 bar) and 3 L (at 2 bar) hold the same gas at the same temperature. Final pressure when they are connected?Answer: $p = \frac{2 \times 1 + 3 \times 2}{5} = 1.6$ bar.

Common Mistakes to Avoid

Watch out

Using $^{\circ}$ C in a gas law. Every ratio of temperatures needs kelvin; $T = t + 273.15$ .
Mixing units of $R$ : $0.0821$ goes with atm and L, $0.08314$ with bar and L, $8.314$ with Pa and $m^{3}$ .
Drawing a Charles's law $V$ vs $t$ ( $^{\circ}$ C) line through the origin. It cuts the $t$ -axis at $- 273.15^{\circ} C$ ; only $V$ vs $T$ (K) passes through the origin.
Using the combined gas law when gas leaks or is added. If $n$ changes, keep $n$ in the equation.
Taking partial pressures in the ratio of masses. Partial pressure follows mole fraction.
Forgetting to subtract aqueous tension for a gas collected over water.
Using Graham's rate ratio as a time ratio. Rate $\propto 1/ M$ , but time for equal volumes $\propto M$ .
Mixing the two STPs: 22.7 L mol $^{- 1}$ at 1 bar, 22.4 L mol $^{- 1}$ at 1 atm.

Frequently Asked Questions

What are the gas laws in chemistry?

The gas laws relate pressure, volume, temperature and amount of a gas. Boyle's law says pV is constant at fixed T, Charles's law says V/T is constant at fixed p, Gay-Lussac's law says p/T is constant at fixed V, and Avogadro's law says V is proportional to n. Together they give $p V = n R T$ . Dalton's and Graham's laws cover mixtures and diffusion.

Why must temperature be in kelvin in gas law calculations?

Gas volume and pressure are proportional to absolute temperature, not to Celsius temperature. On the Celsius scale zero is arbitrary, so a ratio such as 54/27 means nothing physical. On the Kelvin scale zero is the true lowest temperature, so a ratio like 327/300 correctly gives the change in volume or pressure.

What is the value of the gas constant R in different units?

R is 8.314 J per kelvin per mole, which equals 8.314 pascal cubic metre per kelvin per mole. In litre-bar units it is 0.08314, in litre-atmosphere units 0.0821, and in calories 1.987 per kelvin per mole. The value never changes; choose the units that match the pressure and volume in the problem.

What is the molar volume of a gas at STP and SATP?

At the current IUPAC STP of 273.15 K and 1 bar, one mole of ideal gas occupies 22.71 L. At the older STP of 273.15 K and 1 atm it occupies 22.41 L. At SATP, 298.15 K and 1 bar, it occupies 24.79 L. Always use the definition the question gives.

How do you find the molar mass of a gas from its density?

Rearranging the ideal gas equation gives d = pM/RT, so M = dRT/p. Measure the density at a known pressure and temperature and substitute. For example, a gas of density 1.25 g per litre at 1 atm and 273 K has M = 1.25 × 0.0821 × 273 / 1 = 28 g/mol.

What is aqueous tension and why is it subtracted?

Aqueous tension is the vapour pressure of water at the given temperature. A gas collected over water is saturated with water vapour, so the measured pressure is the sum of the dry gas pressure and the aqueous tension. Subtracting it gives the pressure of the dry gas used in calculations.

Are gas laws in the JEE Advanced 2026 syllabus?

Yes. JEE Advanced lists States of Matter: Gases and Liquids, which covers gas laws, the ideal gas equation, the absolute temperature scale, the law of partial pressures and diffusion of gases. The chapter was removed from the rationalised NCERT Class 11 book and from the JEE Main and NEET syllabi, so it is examined mainly in JEE Advanced.

What kind of gas law questions come in JEE Advanced?

JEE Advanced favours graph identification (isotherms, isobars, log plots), connected vessels at different temperatures, partial pressures from masses, gas collected over water, balloons and pay load, and Graham's law with unequal pressures or times. Many are integer-type answers that need careful unit conversion rather than long algebra.

Previous year questions on Gas Laws

2 questions from past papers, each with a step-by-step solution.

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