JEE Mains Physics · Kinetic Theory

Internal Energy and Gas Mixtures

The internal energy of an ideal gas is n(f/2)RT, all of it kinetic; a mixture behaves like one gas whose degrees of freedom and Cv are mole-weighted averages, and gases mixed at different temperatures share their total energy.

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Why this matters

Eighteen PYQs, four of them asking for a number, and one from 2026. Ten find an internal energy or the heat needed to change it, and eight replace a mixture by one equivalent gas or find the temperature after mixing. Weight everything by moles, and average f or Cv, never γ.

Concept 1 of 2: Internal energy U = n(f/2)RT

The molecules of an ideal gas do not attract each other, so there is no potential energy between them: the internal energy is all kinetic, (f/2)kT per molecule. It depends on the temperature alone. Because nRT = PV, the internal energy can also be found from the pressure and volume without knowing T.

Definition

$U = n\dfrac{f}{2}RT = N\dfrac{f}{2}kT = \dfrac{f}{2}PV$ .
$\Delta U = nC_v\Delta T$ for any process; at constant volume the heat supplied is exactly this: $Q = nC_v\Delta T$ .
The translational part alone is $\dfrac{3}{2}nRT$ .
To multiply $v_{rms}$ by a factor k, multiply the kelvin temperature by $k^{2}$ .
A mixture: add the internal energies of its parts, $U = \sum n_i\dfrac{f_i}{2}RT$ .
An insulated container of gas moving at speed v and stopped suddenly: the ordered kinetic energy, $\tfrac{1}{2}Mv^{2}$ per mole, becomes internal energy, so $C_v\Delta T = \tfrac{1}{2}Mv^{2}$ .

Internal energy of an ideal gas

U = n\frac{f}{2}RT = \frac{f}{2}PV \qquad \Delta U = nC_v\Delta T

Worked example

A vessel holds 3 mol of neon and 2 mol of nitrogen at temperature T. Ignoring vibration, find the total internal energy, and its value at 300 K.

(R = 8.31)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q98Moderate

Example 1 · Kinetic Theory · Internal Energy and Gas Mixtures

A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

k for molecules, R for moles

Ten molecules carry an energy measured in kT; ten moles carry one measured in RT. The options often offer both with the same number in front.

"Neglect vibration" sets f = 5

A diatomic gas with vibration ignored is rigid: f = 5, not 7. Only an explicit vibrational mode adds 2.

Drill 9 more on internal energy u = n(f/2)rt

Concept 2 of 2: A mixture as one equivalent gas, and the temperature after mixing

In a mixture the energies add, so the heat capacity of the whole is the sum of the parts. Divide by the total moles and you get the mixture's Cv: a mole-weighted mean. The same holds for f, and γ then follows from 1 + 2/f. When gases at different temperatures are mixed with no energy lost, the total internal energy before equals the total after, which fixes the final temperature.

Definition

$f_{mix} = \dfrac{n_1f_1 + n_2f_2}{n_1 + n_2}$ , $C_{v,mix} = \dfrac{n_1C_{v1} + n_2C_{v2}}{n_1 + n_2} = \dfrac{f_{mix}}{2}R$ .
$C_{p,mix} = C_{v,mix} + R$ , $\gamma_{mix} = 1 + \dfrac{2}{f_{mix}}$ . Equivalently $\dfrac{n_1 + n_2}{\gamma_{mix} - 1} = \dfrac{n_1}{\gamma_1 - 1} + \dfrac{n_2}{\gamma_2 - 1}$ .
Mixing at different temperatures with no loss: $T = \dfrac{\sum n_if_iT_i}{\sum n_if_i}$ . If all the f are equal, this is the mole-weighted mean of the temperatures.
Speed of sound $v_s = \sqrt{\gamma RT/M}$ and $v_{rms} = \sqrt{3RT/M}$ , so $\dfrac{v_{rms}}{v_s} = \sqrt{\dfrac{3}{\gamma}}$ . In a mixture use $\gamma_{mix}$ and the mole-weighted molar mass.

Equivalent gas

f_{mix} = \frac{n_1f_1 + n_2f_2}{n_1 + n_2} \qquad \gamma_{mix} = 1 + \frac{2}{f_{mix}}

Worked example

2 mol of helium are mixed with 3 mol of a rigid diatomic gas. Find

C_v

C_p

and

\gamma

for the mixture.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q108Moderate

Example 2 · Kinetic Theory · Internal Energy and Gas Mixtures

N moles of a polyatomic gas

(f = 6

) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of

N

is:

Averaging γ directly

For equal moles of a monatomic and a rigid diatomic gas, the mean of 5/3 and 7/5 is about 1.53, but the true γ is 1.5. Average f or Cv first, then find γ.

Weighting temperatures by moles alone

When the gases have different f, each one's energy is n(f/2)RT, so the final temperature is weighted by nf, not by n.

Drill 7 more on a mixture as one equivalent gas, and the temperature after mixing

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 (2)

Internal energy U = n(f/2)RT
Internal energy of an ideal gas
$U = n\frac{f}{2}RT = \frac{f}{2}PV \qquad \Delta U = nC_v\Delta T$
A mixture as one equivalent gas, and the temperature after mixing
Equivalent gas
$f_{mix} = \frac{n_1f_1 + n_2f_2}{n_1 + n_2} \qquad \gamma_{mix} = 1 + \frac{2}{f_{mix}}$

Watch out for (4)

k for molecules, R for moles
→ Internal energy U = n(f/2)RT
"Neglect vibration" sets f = 5
→ Internal energy U = n(f/2)RT
Averaging γ directly
→ A mixture as one equivalent gas, and the temperature after mixing
Weighting temperatures by moles alone
→ A mixture as one equivalent gas, and the temperature after mixing

Test yourself on Kinetic Theory

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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