Ideal Gases — Question 9

Question 9

A compressibility factor tests the ideal approximation

Use $R=8.314\,\mathrm{J/(mol\,K)}$; do not assume the measured gas is ideal. A gas sample has $n=1.00$ mol, $T=300$ K, measured pressure 1.00 MPa and volume .00270 m$^3$. Define compressibility factor $Z=pV/(nRT)$. A second measurement at the same temperature and amount has $p=.100$ MPa and $V=.0250$ m$^3$.

1. Find $Z$ at both pressures and the ideal-gas volume at 1.00 MPa.

2. For the first state, quantify the error in the amount inferred by incorrectly imposing $Z=1$. Can two measurements alone establish a universal real-gas equation of state?

Question 9 - Solution

A compressibility factor tests the ideal approximation

(a) Method and calculation

$$Z_1=\frac{10^6(.00270)}{8.314(300)}\approx1.08251,\quad Z_2=\frac{10^5(.0250)}{8.314(300)}\approx1.00233.$$$$V_{\mathrm{ideal,1}}=\frac{8.314(300)}{10^6}=.00249420\,\mathrm{m^3}.$$

(b) Interpretation and further calculation

$$n_{\mathrm{inferred}}=\frac{pV}{RT}=Z_1n\approx1.08251\,\mathrm{mol}.$$$$\frac{n_{\mathrm{inferred}}-n}{n}=Z_1-1\approx8.2511\%.$$The low-pressure point is closer to ideal behavior, but two values do not determine temperature dependence or a unique constitutive law.

Physical check

$Z$ is dimensionless and equals one for an ideal gas; its numerical proximity to one must be judged against the needed accuracy.