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 valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be:

T_{1} + T_{2}

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(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 $\frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{2})}{P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1}}$
As there is no loss of energy the initial energy is going to be equal to final energy , average kinetic energy per molecule is $\frac{3}{2} K T$

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

initial KE = $\frac{3}{2} n_{1} K T_{1} + \frac{3}{2} n_{2} K T_{2}$

final KE = $\frac{3}{2} (n_{1} + n_{2}) K T$

So intial KE = final KE

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

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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