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Thermodynamics · Part 4 of 10

Molar Specific Heats: Cv, Cp and γ

A gas's specific heat depends on how it is heated. This part derives the two standard ones from the first law, $C_{v} = \frac{f}{2} R$ at constant volume and $C_{p} = C_{v} + R$ at constant pressure (Mayer's relation), then their ratio γ for one gas and for a mixture.

Builds on: Part 2 · First Law of Thermodynamics.

Video coming soon

Constant volume

$C_{v} = \frac{f}{2} R$

With the volume fixed, $d W = 0$ , so $d Q = d U$ : $n C_{v} d T = \frac{f}{2} n R d T$ , giving $C_{v} = \frac{f}{2} R$ ( $\frac{3}{2} R$ monatomic, $\frac{5}{2} R$ diatomic).

A rigid box of gas heated by a flame. — Locked lid: all the heat raises U.

A gas under a free piston heated by a flame. — Free piston: some heat goes into work.

Constant pressure

Mayer: $C_{p} = C_{v} + R$

At constant pressure the gas also does work $P d V = n R d T$ . So $n C_{p} d T = n C_{v} d T + n R d T$ and $C_{p} = C_{v} + R = \frac{f + 2}{2} R$ : always bigger than $C_{v}$ .

Ratio of heat capacities

$γ = \frac{C _{p}}{C _{v}} = \frac{f + 2}{f}$

Monatomic 5/3 ≈ 1.67, diatomic 7/5 = 1.4, non-linear polyatomic (f = 6) 4/3. More degrees of freedom, smaller γ; always γ > 1.

Table of f, C_v, C_p and gamma. — C_v, C_p and γ for common gases.

Derivation of gamma for a gas mixture. — Average C_v, not γ.

Mixtures

$\sum \frac{n _{i}}{γ _{i} - 1} = \frac{\sum n _{i}}{γ _{mi x} - 1}$

Internal energies add at the common temperature, so $C_{v}$ values average by moles. With $C_{v} = R / (γ - 1)$ this gives the rule above. Never average γ directly.

Summary

Key formulas

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

Molar Heat Capacity at Constant Volume (C_v)

C_{p} = C_{v} + R = \frac{f + 2}{2} R

Molar Heat Capacity at Constant Pressure (C_p)

γ = \frac{C _{p}}{C _{v}} = \frac{f + 2}{f}

Ratio of Heat Capacities of a Gas

\frac{n _{1}}{γ _{1} - 1} + \frac{n _{2}}{γ _{2} - 1} = \frac{n _{1} + n _{2}}{γ _{e q} - 1}

Ratio of Specific Heats for a Mixture of Gases

Worked examples

One for every idea

Molar Heat Capacity at Constant Volume (C_v)

1. Find the heat required to raise the temperature of 1 mol of a diatomic gas by 10 K at constant volume ( $C_{v} = \frac{5}{2} R$ ).

Diatomic: $f = 5$ , $C_{v} = \frac{5}{2} R = 20.785$ J/mol·K.
$Q = n C_{v} Δ T = 1 \times 20.785 \times 10$ .

$Q \approx 207.9$ J.

Molar Heat Capacity at Constant Pressure (C_p)

2. A monatomic gas has $C_{v} = \frac{3}{2} R$ . Using Mayer's relation, find $C_{p}$ .

$C_{p} = C_{v} + R = \frac{3}{2} R + R$ .

$C_{p} = \frac{5}{2} R \approx 20.79$ J/mol·K.

Ratio of Heat Capacities of a Gas

3. Find γ for a diatomic gas at ordinary temperature.

Vibration is frozen: 3 translational + 2 rotational, $f = 5$ .
$γ = \frac{f + 2}{f} = \frac{7}{5}$ .

$γ = 1.4$ .

Ratio of Specific Heats for a Mixture of Gases

4. A mixture contains 2 mol of a monatomic gas ( $γ_{1} = \frac{5}{3}$ ) and 3 mol of a diatomic gas ( $γ_{2} = \frac{7}{5}$ ). Find the equivalent γ of the mixture.

$\frac{2}{2/3} + \frac{3}{2/5} = 3 + 7.5 = 10.5$ .
$\frac{5}{γ _{mi x} - 1} = 10.5 \Rightarrow γ_{mi x} - 1 = 0.476$ .
A plain mole-weighted average of γ (1.51) is wrong.

$γ_{mi x} \approx 1.48$ .

JEE-style question

Your turn

For an ideal gas, C_p/C_v = 1.4. Its molar heat capacity at constant volume C_v is:

(a)1.5R

(b)2.5R

(c)3.5R

(d)1.4R

Show the answer and the traps

C_v = R/(γ − 1) = R/0.4 = 2.5R: option b.

1.5R is the monatomic value. 3.5R is C_p, not C_v.

And 1.4R confuses γ, a ratio, with C_v/R.

Watch out

Common mistakes

Using C_v in a constant-pressure problem (or the reverse)Constant volume:

C_{v}

. Constant pressure:

C_{p} = C_{v} + R

Averaging γ by moles for a mixtureAverage

C_{v}

(equivalently add

n / (γ - 1)

), then find γ.

Using f = 7 for a diatomic gas at room temperatureVibration is frozen at ordinary temperatures: f = 5, γ = 1.4.

Practice

Try these

1. Find

C_{p}

and γ for a gas with

C_{v} = 3 R

$C_{p} = 4 R$ , $γ = 4/3$ .

2. For a gas,

C_{p} = 29.1

J/mol·K. Find

C_{v}

$C_{v} = C_{p} - R \approx 20.8$ J/mol·K (a diatomic gas).

3. A gas has γ = 1.4. Find its degrees of freedom.

$\frac{f + 2}{f} = 1.4 \Rightarrow f = 5$ .

4. 1 mol of He (γ = 5/3) is mixed with 1 mol of O₂ (γ = 7/5). Find γ of the mixture.

$\frac{1}{2/3} + \frac{1}{2/5} = 4 = \frac{2}{γ - 1} \Rightarrow γ = 1.5$ .