One Mole Of Ideal Gas Is Allowed To Expand Reversibly and adiabatically from a temperature of 27°C. If The Work Done By The Gas Is In The Process Is 3kJ, The Final Temperature Will Be Equal To? (Cv = 20J/K Mol)
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Answer:
∆U=q+w.
Explanation:
For adiabatic process, q=0. So, ∆U=W.Now, Cv=dU/dT. So, dU=Cv*dT and hence W=Cv*dT. (2)Now, W=3kJ=3000J.So, Change in temperature=dT= W/Cv = 150 K.Initial temperature=27°C=300 K.So, Final temperature= 300-150 (since,dT=150) = 150 K.
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Answer is B.∆U=q+w. (1)For adiabatic process, q=0. So, ∆U=W.Now, Cv=dU/dT. So, dU=Cv*dT and hence W=Cv*dT. (2)Now, W=3kJ=3000J.So, Change in temperature=dT= W/Cv = 150 K.Initial temperature=27°C=300 K.So, Final temperature= 300-150 (since,dT=150) = 150 K.
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