Degrees of Freedom and Molar Specific Heats — Question 5

Question 5

The heat-capacity ratio of a mixture is not the average of species ratios

Use classical equipartition: each active independent quadratic energy term contributes $\tfrac12k_BT$ per molecule. Let $f$ count these terms, so for constant $f$, $C_V=fR/2$, $C_p=C_V+R$ and $\gamma=C_p/C_V$. Take $R=8.314$ J/(mol K). An ideal nonreacting mixture contains equal mole amounts of monatomic species A with $f_A=3$ and rigid nonlinear species B with $f_B=6$. Both have the same equilibrium temperature.

1. Find mixture $C_V$, $C_p$ and $\gamma$.

2. Compare with the arithmetic mean of the two pure-species values of $\gamma$. For general mole fraction $x$ of B, derive $\gamma(x)$ and determine whether it increases or decreases as B is added.

Question 5 - Solution

The heat-capacity ratio of a mixture is not the average of species ratios

(a) Method and calculation

$$C_{V,\mathrm{mix}}=\tfrac12(\tfrac32R)+\tfrac12(3R)=\tfrac94R,\quad C_{p,\mathrm{mix}}=\tfrac{13}4R.$$$$\gamma_{\mathrm{mix}}=13/9\approx1.44444,\quad\tfrac12(5/3+4/3)=1.500.$$

(b) Interpretation and further calculation

$$f(x)=3+3x,\quad\gamma(x)=1+\frac2{3+3x}.$$$$\frac{d\gamma}{dx}=-\frac6{(3+3x)^2}<0.$$Capacities add before their ratio is taken. Additional internal modes increase storage capacity relative to the fixed $R$ expansion contribution.

Physical check

The mixture ratio lies between the component ratios but need not equal their mole-weighted average.