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

$T_{1} + T_{2}$

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$\frac{T_{1} + T_{2}}{2}$

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

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

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Solution

The correct option is C
$\frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{2})}{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1})}$

According to the kinetic theory, the average kinetic energy(KE) per molecule of a gas

= $\frac{3}{2} K T$ . 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), $E_{2} = E_{1}$ or

$\frac{3}{2} (n_{1} + n_{2}) k T = \frac{3}{2} k (n_{1} T_{1} + n_{2} T_{2})$

or $T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}$

Now $P_{1} V_{1} = n_{1} R T_{1}$ and $P_{2} V_{2} = n_{2} R T_{2}$ which give

$n_{1} = \frac{P_{1} V_{1}}{R T_{1}}$ and $n_{2} = \frac{P_{2} V_{2}}{R T_{2}}$

Using these in Eq. (1) and simplifying, we get

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

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