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Question

# A box contains N molecules of a perfect ideal gas at temperature T1 and pressure P1. 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 P2 and new temperature is T2, then

A
P2=P1,T2=T1
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B
P2=P1,T2=T12
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C
P2=2P1,T2=T1
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D
P2=2P1,T2=T12
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Solution

## The correct option is B P2=P1,T2=T12Given: For initial case, Number of molecules of gas, N1=N Temperature =T1 Pressure =P1 For final case, Number of molecules of gas, N2=2N Temperature, T2=? Pressure, P2=? Volume (V) of the container will be same in both cases. We know that, average translational kinetic energy is given by, KE=PVf2 ........(1) and KE=NfKBT2 ........(2) where, N→ number of molecules f→ no of degrees of freedom [constant] KB→ Boltzmann constant [constant] T→ temperature As volume of container and average translational kinetic energy has not been changed, so from (1), pressure will not change. ⇒P2=P1 Also from (2), T∝1N Therefore, temperature will be halved if the number of molecules is doubled. ⇒T2=T12

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