A vessel contains N molecules of a gas at temperature T. Now the number of molecules is doubled, keeping the total energy in the vessel constant. The temperature of the gas is.
A
T
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B
2T
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C
T2
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D
√2T
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Solution
The correct option is CT2 We know that, E=32nRT Now, E1=32nRT1...(i) E2=322nRT2...(ii) From equation (i) and (ii) 2E1=E2 Dividing equation (i) and (ii) E1=32nRT1E2=322nRT2 E1E2=T12T2 Energy remainig constant in the vessel, 2=T1T2 Or T2=T12 So, if the temperature taken is T instead of T1 Then, the final temperature becomes T2