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Question

One mole of an ideal Gas (Cv=32R) is heated at constant pressure reversibly of 1 atmosphere from 25 C to 100 C. Calculate the absolute value of ΔE in calories.
(Take R = 2 cal/molK)

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Solution

As, (CpCv)=RCp=Cv+R
Cp=32R+R=52R
Heat given at constant pressure
ΔH=nCpΔT=1×52R(373298)
ΔH=1×52×2×75=375 cal
Work done in the process =PΔV
=P(V2V1)
=P(nRT2PnRT1P)
(As, PV = nRT because the process is carried out reversibly)
=nR(T2T1)
=1×2(373298)
=150 cal
From first law of Thermodynamics,
ΔE=q+W
=375150=225 cal

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