Question

Two thermally insulated vessels 1 and 2 are filled with air at temperatures $(T_{1},T_{2}),$ volume $(V_{1},V_{2})$ and pressure $(P_{1},P_{2})$ respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

• A. $T_{1} + T_{2}$
• B. $(T_{1} + T_{2})/2$
• C. $\frac{T_{1}T_{2}(P_{1}V_{1} + P_{2}V_{2})}{P_{1}V_{1}T_{2} + P_{2}V_{2}T_{1}}$
• D. $\frac{T_{1}T_{2}(P_{1}V_{1} + P_{2}V_{2})}{P_{1}V_{1}T_{1} + P_{2}V_{2}T_{2}}$

Answer 1

Answer: (C)

$\frac{T_{1}T_{2}(P_{1}V_{1} + P_{2}V_{2})}{P_{1}V_{1}T_{2} + P_{2}V_{2}T_{1}}$

Answer 2

The number of moles of the system remains same,

$\frac{P_{1}V_{1}}{RT_{1}} + \frac{P_{2}V_{2}}{RT_{2}} = \frac{P(V_{1} + V_{2})}{RT}$ Ž $T = \frac{P(V_{1} + V_{2})T_{1}T_{2}}{(P_{1}V_{1}T_{2} + P_{2}V_{2}T_{1})}$

According to Boyle’s law,

$P_{1}V_{1} + P_{2}V_{2} = P(V_{1} + V_{2})$ \ $T = \frac{(P_{1}V_{1} + P_{2}V_{2})T_{1}T_{2}}{(P_{1}V_{1}T_{2} + P_{2}V_{2}T_{1})}$