Two thermally insulted vessels $1$ and $2$ are filled with air at temperature $(T_{1}, T_{2})$ , volumes $(V_{1}, V_{2})$ and pressures $(p_{1}, p_{2})$ respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be (P $=$ common pressure).

T_{1} + T_{2}

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(T_{1} + T_{2}) / 2

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\frac{T_{1} T_{2} p (V_{1} + 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_{2} + p_{2} V_{2} T_{1}}

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Solution

The correct option is D $\frac{T_{1} T_{2} p (V_{1} + V_{2})}{p_{1} V_{1} T_{2} + p_{2} V_{2} T_{1}}$
From gas equation we get PV=nRT
Since the number of molecules of gas will remain constant
$\frac{P_{1} V_{1}}{R T_{1}} + \frac{P_{2} V_{2}}{R T_{2}} = \frac{P (V_{1} + V_{2})}{R T} \frac{P (V_{1} + V_{2})}{R T} = \frac{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1}}{R T_{2} T_{1}} T = \frac{P (V_{1} + V_{2}) T_{2} T_{1}}{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1}}$

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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

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