MHT-CET Chemistry · Some Basic Concepts of Chemistry

Real Gases, Dalton's Law and the Kinetic Theory of Gases

In a gas mixture each component pushes independently, so its partial pressure is just its share of the moles times the total pressure; the kinetic theory explains this, gives the speed of the molecules, and shows why real gases stray from ideal behaviour.

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Why this matters

Thirteen PYQs, and more than half are a single trick: partial pressure is proportional to moles, so equal masses of two gases do NOT share pressure equally. The rest split between one root-mean-square-velocity ratio, the compressibility factor Z as the measure of non-ideality, and one recall question on liquefaction. Master 'moles first, never grams' and you have the whole subtopic — the arithmetic is easy once the mole fractions are right.

Concept 1 of 4

Dalton's law of partial pressures

Intuition

In a mixture, each gas ignores the others and pushes on the walls as if it were alone; the total pressure is just the sum of those individual pushes. Because pressure at fixed temperature and volume depends only on how many molecules there are, each gas's share of the pressure equals its share of the moles.

Definition

Dalton's law and the mole-fraction rule:

The total pressure of a non-reacting gas mixture equals the sum of the partial pressures of its components: $P_{\text{total}} = P_1 + P_2 + \cdots$ .
The partial pressure of a component is its mole fraction times the total pressure: $P_i = x_i\,P_{\text{total}}$ .
Mole fraction $x_i = \dfrac{n_i}{n_{\text{total}}}$ , so at fixed $T$ and $V$ , partial pressure is proportional to moles.
Equal-mass shortcut: if the components have equal masses, then $n \propto \dfrac{1}{M}$ — the lighter gas has more moles and therefore the larger partial pressure.

Partial pressure from mole fraction

P_i = x_i\,P_{\text{total}} = \dfrac{n_i}{n_{\text{total}}}\,P_{\text{total}}

P_ipartial pressure of component i
x_imole fraction of component i
n_imoles of component i
P_{\text{total}}total pressure of the mixture

Worked example

A vessel holds 16 g of O2, 14 g of N2 and 2 g of H2 at the same temperature. If the total pressure is 12 bar, find the partial pressure of O2. (O2 = 32, N2 = 28, H2 = 2)

Convert each mass to moles: $n_{\text{O}_2} = 16/32 = 0.5$ , $n_{\text{N}_2} = 14/28 = 0.5$ , $n_{\text{H}_2} = 2/2 = 1$ .
Total moles $= 0.5 + 0.5 + 1 = 2$ mol.
Mole fraction of $\text{O}_2$ : $x_{\text{O}_2} = 0.5/2 = 0.25$ .
Partial pressure: $P_{\text{O}_2} = x_{\text{O}_2}\,P_{\text{total}} = 0.25 \times 12 = 3$ bar.

Answer:

P_{\text{O}_2} = 3

bar.

Practice this conceptself-check · 3 quick reps

Try it yourself

Equal masses of helium and oxygen are mixed in a container. What fraction of the total pressure is exerted by helium? (He = 4, O2 = 32)

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
In a mixture, a gas has mole fraction 0.4 and the total pressure is 5 bar. Its partial pressure?
2.
Equal masses of H2 (M = 2) and He (M = 4) are mixed. Ratio of their partial pressures H2 : He?
3.
A mixture has 1 mol N2 and 4 mol He. Mole fraction of N2?

From the bank · past-year question

Example 1States of MatterMODERATE

A container consisting mixture of

28\text{ }g{\text{ }N}_{2},8\text{ }gHe

and 40 g Ne at

25^{\circ}C

. If the total pressure exerted by the gaseous mixture is 20 bar, what is the partial pressure exerted by

N_{2}

[Q61 · 20 April Shift II · 2025]

Partial pressure follows moles, not mass

Equal masses of two gases do NOT exert equal partial pressures. Convert every mass to moles first: the lighter gas (smaller

M

) has more moles and the larger partial pressure. For equal masses of

\text{H}_2

and He, the ratio is

2:1

, not

1:1

Use the total moles in the denominator

Mole fraction is

x_i = n_i / n_{\text{total}}

, where

n_{\text{total}}

is the sum over all gases in the mixture. Forgetting one component inflates every mole fraction and the sum of all

x_i

will not equal 1.

Concept 2 of 4

Root-mean-square velocity

Intuition

The molecules of a gas move at a spread of speeds; the root-mean-square velocity is the effective average speed that carries the kinetic energy. It rises with temperature (hotter means faster) and falls with molar mass (heavier molecules are more sluggish).

Definition

Root-mean-square (rms) velocity:

$v_{rms} = \sqrt{\dfrac{3RT}{M}}$ — it grows as $\sqrt{T}$ and shrinks as $1/\sqrt{M}$ .
To compare two gases (or the same gas at two states), take the ratio and cancel the constant $3R$ : $\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1/M_1}{T_2/M_2}}$ .
The square root is essential — a ratio of $4$ inside the root becomes $2$ outside it.

Root-mean-square velocity and its ratio

v_{rms} = \sqrt{\dfrac{3RT}{M}}\qquad \dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1/M_1}{T_2/M_2}}

v_{rms}root-mean-square velocity
Runiversal gas constant
Tabsolute temperature (K)
Mmolar mass (kg/mol in SI)

Worked example

Find the ratio of the rms velocity of H2 at 50 K to that of O2 at 800 K. (H2 = 2, O2 = 32)

Write the ratio: $\dfrac{v_{\text{H}_2}}{v_{\text{O}_2}} = \sqrt{\dfrac{T_{\text{H}_2}/M_{\text{H}_2}}{T_{\text{O}_2}/M_{\text{O}_2}}}$ .
Substitute: $= \sqrt{\dfrac{50/2}{800/32}} = \sqrt{\dfrac{25}{25}}$ .
The fraction inside is $1$ , and $\sqrt{1} = 1$ .

Answer:The ratio is

1

(the two rms velocities are equal).

Practice this conceptself-check · 3 quick reps

Try it yourself

At the same temperature, what is the ratio of the rms velocity of H2 to that of O2? (H2 = 2, O2 = 32)

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
At the same temperature, which is faster on average — H2 or O2?
2.
If the absolute temperature of a gas is made 4 times larger, by what factor does v_rms change?
3.
For the same gas, ratio of v_rms at 400 K to that at 100 K?

From the bank · past-year question

Example 2States of MatterMODERATE

The ratio between the root mean square velocity of H

_2

at 50 K and that of O

_2

at 800 K is

[Q58 · May Shift 1 · 2021]

Take the square root at the end

The molar-mass and temperature factors sit inside a square root. If the combined ratio inside works out to

4

, the velocity ratio is

\sqrt{4} = 2

, not

4

. Forgetting the root is the most common error in these ratio problems.

Use absolute temperature in kelvin

v_{rms} = \sqrt{3RT/M}

needs

T

in kelvin. Never plug in Celsius — convert

t\,^\circ\text{C}

T = t + 273

K first.

Concept 3 of 4

Postulates of the kinetic theory of gases

Intuition

The kinetic theory pictures a gas as a huge number of tiny, fast, independent particles bouncing around in empty space. Its handful of assumptions are exactly what make the ideal gas laws work — and knowing which one fails tells you why real gases misbehave.

Definition

The kinetic theory of gases (KTG) rests on a few idealising assumptions:

Gas molecules are point masses — their own volume is negligible compared with the container.
There are no attractive or repulsive forces between molecules.
Collisions are perfectly elastic, so no kinetic energy is lost.
The average kinetic energy is proportional to the absolute temperature.

The two assumptions in bold above (zero molecular volume, zero intermolecular force) are precisely the ones that break down for a real gas.

Postulate	Statement
Negligible molecular volume	The actual volume of the gas molecules is negligibly small compared with the total volume of the container; the gas is mostly empty space. This assumption fails at high pressure, when molecules are squeezed close together and their own volume is no longer negligible.
No intermolecular forces	There are no forces of attraction or repulsion between the molecules of an ideal gas; they move completely independently. This assumption fails at low temperature / high pressure, when attractions pull molecules together — the reason gases can be liquefied.
Elastic collisions	Collisions between molecules, and with the walls, are perfectly elastic — the total kinetic energy is conserved during every collision.
Kinetic energy proportional to temperature	The average kinetic energy of the molecules is directly proportional to the absolute temperature; it depends only on T, not on the gas's identity.
Continuous random motion	Molecules are in constant, rapid, random straight-line motion in all directions, colliding with one another and the container walls.

The two bold postulates (zero volume, zero force) are what an ideal gas assumes and a real gas violates.

Practice this conceptself-check · 4 quick reps

Try it yourself

Among SO2, Cl2, NH3 and O2, which gas is the most difficult to liquefy, and which kinetic-theory assumption explains why gases can be liquefied at all?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
In the kinetic theory, the volume of the gas molecules themselves is assumed to be what?
2.
According to KTG, molecular collisions are of what type?
3.
The average kinetic energy of gas molecules is proportional to what?
4.
Which KTG assumption must fail for a gas to be liquefiable?

From the bank · past-year question

Example 3States of MatterEASY

Which among the following gases is difficult to liquify?

[Q78 · 15th May Shift 2 · 2023]

Kinetic energy depends on temperature, not on the gas

The average kinetic energy of gas molecules depends only on the absolute temperature — at the same

T

\text{H}_2

and

\text{O}_2

have the same average kinetic energy. They differ in *speed* (the lighter gas moves faster), not in energy.

Ideal gas = zero volume AND zero force

An ideal gas assumes both that molecules have no volume and that there are no forces between them. A real gas violates both, which is why it deviates most where these matter — at high pressure and low temperature.

Concept 4 of 4

Real gases and the compressibility factor

Intuition

A real gas obeys

PV = nRT

only approximately, because its molecules do take up space and do attract one another. The compressibility factor Z measures how far the gas strays from ideal: Z = 1 is perfectly ideal, and any departure from 1 signals real-gas behaviour.

Definition

Deviation from ideal behaviour:

The compressibility factor $Z = \dfrac{PV}{nRT}$ . For an ideal gas $Z = 1$ at all conditions; for a real gas $Z \ne 1$ .
Since $Z = V_{\text{real}}/V_{\text{ideal}}$ , the real molar volume is $V_{\text{real}} = Z \times V_{\text{ideal}}$ (at STP $V_{\text{ideal}} = 22.4\,\text{dm}^3$ ).
The van der Waals equation corrects both flaws: $\left(P + \dfrac{an^2}{V^2}\right)(V - nb) = nRT$ , where $a$ accounts for intermolecular attraction and $b$ for the finite volume of the molecules.
Deviations are greatest at high pressure and low temperature; gases with stronger attractions (higher $a$ , higher critical temperature) are easier to liquefy.

Compressibility factor and van der Waals equation

Z = \dfrac{PV}{nRT}\qquad \left(P + \dfrac{an^2}{V^2}\right)(V - nb) = nRT

Zcompressibility factor (1 for ideal, ≠ 1 for real)
avan der Waals constant for intermolecular attraction
bvan der Waals constant for molecular volume

Worked example

The compressibility factor of a real gas is 1.05 at STP. What is its molar volume? (Ideal molar volume at STP = 22.4 dm3)

Use $Z = \dfrac{V_{\text{real}}}{V_{\text{ideal}}}$ , so $V_{\text{real}} = Z \times V_{\text{ideal}}$ .
Substitute: $V_{\text{real}} = 1.05 \times 22.4$ .
$1.05 \times 22.4 = 23.52\,\text{dm}^3$ .

Answer:

V_{\text{real}} = 23.52\,\text{dm}^3

Practice this conceptself-check · 4 quick reps

Try it yourself

A real gas has compressibility factor 1.1 at STP. Find its molar volume. (Ideal molar volume at STP = 22.4 dm3)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
What is the compressibility factor Z of an ideal gas?
2.
Write the formula for the compressibility factor.
3.
In the van der Waals equation, which constant corrects for the volume of the molecules?
4.
Deviations from ideal behaviour are largest at what conditions?

From the bank · past-year question

Example 4States of MatterEASY

If compressibility factor of real gas is 1.05 at STP. What is molar volume of real gas?

[Q75 · 20 April Shift I · 2025]

Z = 1 means ideal, in either direction

A gas is ideal only when

Z = 1

. Both

Z > 1

(repulsion / molecular volume dominates) and

Z < 1

(attraction dominates) mean the gas is real — the deviation, not its sign, is what matters.

Multiply, do not add, the ideal molar volume

V_{\text{real}} = Z \times V_{\text{ideal}}

. For

Z = 1.05

at STP that is

1.05 \times 22.4 = 23.52\,\text{dm}^3

, not

22.4 + 1.05

. Z is a multiplying factor, not an offset.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Dalton's law of partial pressures
Partial pressure from mole fraction
$P_i = x_i\,P_{\text{total}} = \dfrac{n_i}{n_{\text{total}}}\,P_{\text{total}}$
Root-mean-square velocity
Root-mean-square velocity and its ratio
$v_{rms} = \sqrt{\dfrac{3RT}{M}}\qquad \dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1/M_1}{T_2/M_2}}$
Real gases and the compressibility factor
Compressibility factor and van der Waals equation
$Z = \dfrac{PV}{nRT}\qquad \left(P + \dfrac{an^2}{V^2}\right)(V - nb) = nRT$

Reference tables (1)

Postulates of the kinetic theory of gases5 rows

Postulate	Statement
Negligible molecular volume	The actual volume of the gas molecules is negligibly small compared with the total volume of the container; the gas is mostly empty space. This assumption fails at high pressure, when molecules are squeezed close together and their own volume is no longer negligible.
No intermolecular forces	There are no forces of attraction or repulsion between the molecules of an ideal gas; they move completely independently. This assumption fails at low temperature / high pressure, when attractions pull molecules together — the reason gases can be liquefied.
Elastic collisions	Collisions between molecules, and with the walls, are perfectly elastic — the total kinetic energy is conserved during every collision.
Kinetic energy proportional to temperature	The average kinetic energy of the molecules is directly proportional to the absolute temperature; it depends only on T, not on the gas's identity.
Continuous random motion	Molecules are in constant, rapid, random straight-line motion in all directions, colliding with one another and the container walls.

The two bold postulates (zero volume, zero force) are what an ideal gas assumes and a real gas violates.

Watch out for (8)

Partial pressure follows moles, not mass
→ Dalton's law of partial pressures
Use the total moles in the denominator
→ Dalton's law of partial pressures
Take the square root at the end
→ Root-mean-square velocity
Use absolute temperature in kelvin
→ Root-mean-square velocity
Kinetic energy depends on temperature, not on the gas
→ Postulates of the kinetic theory of gases
Ideal gas = zero volume AND zero force
→ Postulates of the kinetic theory of gases
Z = 1 means ideal, in either direction
→ Real gases and the compressibility factor
Multiply, do not add, the ideal molar volume
→ Real gases and the compressibility factor

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