The Molar Specific Heats of an Ideal Gas — Question 4

Question 4

A mixture heat capacity can reveal its mole fraction

Use $pV=nRT$, $R=8.314$ J/(mol K), and the sign convention $\Delta U=Q-W$, with $W$ the work done by the gas. Only pressure–volume work occurs. A nonreacting binary ideal mixture has constant species capacities $C_{V,A}=\tfrac32R$ and $C_{V,B}=\tfrac52R$. Its measured molar $C_{p,\mathrm{mix}}=3.20R$. Species masses are $M_A=4.00$ and $M_B=28.0$ g/mol.

1. Find the mole fraction and mass fraction of B.

2. For .500 mol of mixture heated by 100 K at fixed volume, find $Q$ and the ratio $\gamma_{\mathrm{mix}}$. State the capacity range for a physically possible mixture of these two species.

Question 4 - Solution

A mixture heat capacity can reveal its mole fraction

(a) Method and calculation

$$C_{p,\mathrm{mix}}=(1-x_B)(2.50R)+x_B(3.50R)=3.20R\ \Rightarrow\ x_B=.700.$$$$w_B=\frac{x_BM_B}{(1-x_B)M_A+x_BM_B}=\frac{19.6}{20.8}=\frac{49}{52}\approx.942308.$$

(b) Interpretation and further calculation

$$C_{V,\mathrm{mix}}=2.20R,\quad Q=.500(2.20)(8.314)(100)=914.54\,\mathrm J.$$$$\gamma_{\mathrm{mix}}=3.20/2.20=16/11,\quad2.50R\le C_{p,\mathrm{mix}}\le3.50R.$$The weighting is by moles for molar capacity; mass fractions require the separate molar-mass conversion.

Physical check

A mixture capacity outside the component range contradicts this additive constant-composition model.