There are two vessels filled with an ideal gas where the volume of one is double the volume of the other. The larger vessel contains the gas at $8\ \text{kPa}$ at $1000\ \text{K}$ while the smaller vessel contains the gas at $7\ \text{kPa}$ at $500\ \text{K}$. If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at $600\ \text{K}$, at steady state the pressure in the vessels will be (in kPa):
Answer: (B) $6$
The total number of moles is conserved, and at the end both vessels share one pressure $P$ at $600\ \text{K}$. With volumes $2V$ and $V$:
$$\frac{8\times2V}{R\times1000} + \frac{7\times V}{R\times500} = \frac{P\times3V}{R\times600}$$
$$0.016 + 0.014 = \frac{P}{200} \Rightarrow P = 6\ \text{kPa}$$
Solution by Sreeraj P, M.Sc Physics