$N$ molecules of an ideal gas at temperature $T_{1}$ and pressure $P_{1}$ are contained in a closed box. The molecules in the box gets doubled, Keeping total kinetic energy same. If new pressure is $P_{2}$ and temperature is $T_{2}$ , Then:

P_{2} = P, T_{2} = T_{1}

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P_{2} = P_{1}, T_{2} = \frac{T_{1}}{2}

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P_{2} = 2 P_{1} . T_{2} = T_{1}

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P_{2} = 2 P_{1}, T_{2} = \frac{T_{1}}{2}

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Solution

The correct option is B $P_{2} = P_{1}, T_{2} = \frac{T_{1}}{2}$
$K . E = \frac{1}{2} NkT = constant$
$N T = constant \Rightarrow T \propto \frac{1}{N}$
Hence final temperature is halved as $N$ is doubled.
and $K . E = \frac{f}{2} PV = constant$
For a closed container, $V =$ constant. Hence pressure is also constant.

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A box containing N molecules of a perfect gas at temperature $T_{1}$ and pressure $P_{1}$ . The number of molecules in the box is doubled keeping the total kinetic energy of the gas same as before. if the new pressure is $P_{2}$ and temperature $T_{2}$ , then