Three identical adiabatic containers A, B and C contain helium, neon and oxygen respectively at equal pressure. The gases are pushed to half their original volumes.
Since this is an adiabatic process, Q(heat) = 0.
Hence, dU(change in internal energy) = dW(work done) = nCvdT.
where n = no. of moles, Cv = specific heat of the gas at constant volume, and dT = change in temperature.
Cv is different for oxygen(diatomic gas) but same for helium and neon(monoatomic gases) so dT for oxygen is different.
W for adiabatic process = (P2V2-P1V1)/(1-