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Kinetic Theory of Gases · Part 8 of 9

Internal Energy of a Gas

Add up $\frac{1}{2} k T$ for every degree of freedom of every molecule and you get the internal energy of a gas. For an ideal gas it depends on temperature alone, which makes mixing problems quick to solve.

Builds on: Part 7 · Degrees of Freedom and Equipartition of Energy.

Internal Energy of a Gas

Video coming soon

Ideal gas

$U = \frac{f}{2} n R T$

No intermolecular forces means no potential energy: U is all kinetic and depends only on T. Monatomic $\frac{3}{2} n R T$ · diatomic $\frac{5}{2} n R T$ · linear polyatomic $\frac{5 + x}{2} n R T$ · non-linear $\frac{6 + x}{2} n R T$ .

Energy blocks per mole for each type of molecule. — Each block is $\frac{1}{2} R T$ per mole.

Energy blocks of helium and oxygen pooled into a mixture. — Eight shares pooled over two moles: four per mole.

Mixtures

Equivalent degrees of freedom

Add the internal energies: $(f_{1} n_{1} + f_{2} n_{2}) \frac{R T}{2} = f_{eq} (n_{1} + n_{2}) \frac{R T}{2}$ , so $f_{eq} = \frac{\sum f _{i} n _{i}}{\sum n _{i}}$ , a mole-weighted average. 1 mol He + 1 mol O₂: $f_{eq} = \frac{3 + 5}{2} = 4$ .

Insulated mixing

$T_{f} = \frac{\sum f _{i} n _{i} T _{i}}{\sum f _{i} n _{i}}$

In a thermally insulated, rigid vessel, Q = 0 and W = 0, so U before = U after: $\sum f_{i} n_{i} T_{i} = (\sum f_{i} n_{i}) T_{f}$ . For 1 mol He at 600 K and 1 mol N₂ at 200 K: $T_{f} = \frac{3 \times 600 + 5 \times 200}{8} = 350 K$ , not the simple average 400 K.

Energy shares of helium at 600 K and nitrogen at 200 K combined. — N₂ holds more energy per kelvin, so it pulls T_f towards 200 K.

Summary

Key formulas

U = \frac{f}{2} n R T

Internal energy of an ideal gas

monatomic: U = \frac{3}{2} n R T

Internal energy of an ideal gas

diatomic: U = \frac{5}{2} n R T

Internal energy of an ideal gas

linear polyatomic: U = \frac{5 + x}{2} n R T

Internal energy of an ideal gas

non-linear: U = \frac{6 + x}{2} n R T

Internal energy of an ideal gas

f_{eq} = \frac{f _{1} n _{1} + f _{2} n _{2} + \dots}{n _{1} + n _{2} + \dots}

Equivalent degrees of freedom for a mixture

T_{f} = \frac{f _{1} n _{1} T _{1} + f _{2} n _{2} T _{2} + \dots}{f _{1} n _{1} + f _{2} n _{2} + \dots}

Mixing of gases at constant volume

JEE-style question

Your turn

1 mol of He at 600 K mixes with 1 mol of N₂ at 200 K in an insulated rigid vessel. Find the final temperature.

1)350 K

2)400 K

3)450 K

4)366.7 K

Show the answer and the traps

Weight by f n: 350 K. Option 1.

400 K is the simple mole average. 450 K swaps the f values. 366.7 K weights by C_P instead of C_V.

Watch out

Common mistakes

Taking the simple average of temperatures.Weight each gas by f·n: a diatomic stores more energy per kelvin than a monatomic gas.

Weighting by C_P in a rigid vessel.At constant volume the energy per kelvin is

\frac{f}{2} R = C_{V}

.

Practice

Try these

1. Internal energy of 2 mol of O₂ at 27 °C?

$\frac{5}{2} \times 2 \times 8.314 \times 300$ ≈ 1.25 × 10⁴ J.

2. f_eq for 2 mol of He and 1 mol of N₂?

$\frac{2 \times 3 + 1 \times 5}{3} = \frac{11}{3}$ ≈ 3.67.

3. 2 mol of He at 400 K mix with 1 mol of O₂ at 250 K in an insulated rigid vessel. Final temperature?

$\frac{3 \times 2 \times 400 + 5 \times 1 \times 250}{6 + 5} = \frac{3650}{11}$ ≈ 332 K.