Ideal Gases — Question 3

Question 3

Connected bulbs at different temperatures share pressure but not density

Treat the gas as ideal, $pV=nRT$, and use $R=8.314\,\mathrm{J/(mol\,K)}$. Two rigid bulbs of volumes 2.00 L and 3.00 L are joined by a negligible-volume tube. Their temperatures are maintained at 300 K and 450 K. Total gas amount is .400 mol. Neglect gravity and thermal-transpiration effects; assume a common equilibrium mechanical pressure.

1. Find the pressure and gas amount in each bulb.

2. If both bulbs are brought to 300 K without changing total amount or volume, find the final pressure. Does agreement with the arithmetic mean in this example justify that averaging rule in general?

Question 3 - Solution

Connected bulbs at different temperatures share pressure but not density

(a) Method and calculation

$$n=\frac pR\left(\frac{V_1}{T_1}+\frac{V_2}{T_2}\right),\quad p=\frac{.400(8.314)}{.002/300+.003/450}=249420\,\mathrm{Pa}.$$$$n_1=\frac{p(.002)}{R(300)}=.200\,\mathrm{mol},\quad n_2=.200\,\mathrm{mol}.$$

(b) Interpretation and further calculation

$$p_f=\frac{.400(8.314)(300)}{.005}=199536\,\mathrm{Pa}.$$$$T_{\mathrm{effective}}=\frac{V_1+V_2}{V_1/T_1+V_2/T_2}=375\,\mathrm K.$$The required temperature average is volume-weighted harmonic. Here it happens to equal the unweighted arithmetic mean, but the equality follows from these particular volumes and temperatures, not a general rule.

Physical check

At common pressure, number density is proportional to $1/T$; unequal volumes can still hold equal amounts.