Ideal Gases — Question 2

Question 2

A falling pressure can mask simultaneous heating and gas loss

Treat the gas as ideal, $pV=nRT$, and use $R=8.314\,\mathrm{J/(mol\,K)}$. A rigid tank of volume .0200 m$^3$ initially contains gas at 300 kPa and 300 K. After a leak, its pressure is 240 kPa and temperature is 320 K. Pressures are absolute.

1. Find the fraction of gas lost and the amounts before and after.

2. What final pressure would have occurred without leakage at 320 K? Explain why the pressure-drop fraction alone underestimates the lost fraction.

Question 2 - Solution

A falling pressure can mask simultaneous heating and gas loss

(a) Method and calculation

$$\frac{n_f}{n_i}=\frac{p_f/T_f}{p_i/T_i}=\frac{240(300)}{300(320)}=.750.$$$$n_i=\frac{300000(.0200)}{8.314(300)}\approx2.40558\,\mathrm{mol},\quad n_f\approx1.80419\,\mathrm{mol}.$$

(b) Interpretation and further calculation

$$f_{\mathrm{lost}}=1-.750=.250,\quad p_{f,\mathrm{sealed}}=300(320/300)=320\,\mathrm{kPa}.$$$$1-p_f/p_i=.200\ne.250.$$Heating tends to increase pressure, partly concealing the effect of losing molecules.

Physical check

The inferred remaining amount is smaller despite the higher final temperature.