Question

Two thermally insulated vessels 1 and 2 are filled with air at temperatures(T1,T2),volume (V1,V2) and pressure (P1,P2) respectively.If the valve joining the two vessels is opened,the temperature inside the vessel at equilibrium will be?

Answer 1

Answer on Question #42742-Physics-Molecular Physics-Thermodynamics

Two thermally insulated vessels 1 and 2 are filled with air at temperatures (T1, T2), volume (V1, V2) and pressure (P1, P2) respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be?

Solution

According to the kinetic theory, the average kinetic energy (KE) per molecule $f$ a gas is $\frac{3}{2} kT$. Let $n_1$ and $n_2$ be the number of moles of air in vessels 1 and 2 respectively.

Before mixing, the total KE of molecules in the two vessels is

$$E_1 = \frac{3}{2} n_1 k T_1 + \frac{3}{2} n_2 k T_2 = \frac{3}{2} k (n_1 T_1 + n_2 T_2).$$

After mixing, the total KE of molecules is

$$E_2 = \frac{3}{2} (n_1 + n_2) k T,$$

where $T$ is the temperature when equilibrium is established. Since there is no loss of energy (because the vessels are insulated), $E_1 = E_2$ or

$$\frac{3}{2} k (n_1 T_1 + n_2 T_2) = \frac{3}{2} (n_1 + n_2) k T \rightarrow T = \frac{(n_1 T_1 + n_2 T_2)}{(n_1 + n_2)}.$$

Now $P_1 V_1 = n_1 R T_1$ and $P_2 V_2 = n_2 R T_2$ which gives

$$n_1 = \frac{P_1 V_1}{R T_1} \quad \text{and} \quad n_2 = \frac{P_2 V_2}{R T_2}.$$

Using these in equation for $T$ and simplifying, we get

$$T = \frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1}.$$

Answer: $T = \frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1}$.

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