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Question

# Two mole of ideal gas diatomic gas (Cv,m=5/2R) at 300K and 5atm expanded irreversely and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm. Calculate q,w,ΔH and ΔU.

A
q=0,ΔU=w=1247.1J,ΔH=1745.94J
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B
q=0,ΔU=w=+1247.1J,ΔH=1745.94J
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C
U=0,Δq=w=+1247.1J,ΔH=+1745.94J
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D
None of these
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Solution

## The correct option is A q=0,ΔU=w=−1247.1J,ΔH=−1745.94JFor adiabatic change, there is no heat transfer involved. q=0The change in the internal energy is equal to the work done ΔU=wBut ΔU=nCvΔT and w=PΔV=Pavg(nRTfPf−nRTiPi)Hence, nCvΔT=Pavg(nRTfPf−nRTiPi)n×52RΔT=−Pavg(nRTfPf−nRTiPi)5/2(Tf−300)=−(Tf2−3005)5/2Tf−750=−Tf2+603Tf=810 ; Tf=270KThe change in the internal energy isΔU=w=nCVΔT=2×52R(−30)=−150R=−1247.1JThe enthalpy change is ΔH=ΔU+nRΔT=−150R+2R(−30)=−210R=−1745.9J

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