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Question

# Air (ideal gas with γ=1.4) at 1 bar and 300 K is compressed till the final volume is one-sixteenth of the original volume, following a polytropic process PV1.25=const. Calculate the (a) work done and (b) the energy transferred as heat per mole of the air.

A
9.977 kJ/mol,3.742 kJ/mol
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B
6.97 kJ/mol,8.24 kJ/mol
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C
6.97 kJ/mol,2.64 kJ/mol
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D
3.742 kJ/mol, 2.64 kJ/mol
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Solution

## The correct option is A −9.977 kJ/mol,−3.742 kJ/molGiven that gas followsPV1.25=constant Let V1 be volume corresponding to pressure P1=1 bar, V2 be volume corresponding to pressure P2 Also given that V2=V116 Hence, P1V1.251=P2V1.252 P2=P1(V1/V2)1.25=1(16)1.25=32 bar Using P1V1T1=P2V2T2 T2=(T1P2V2)(P1V1)=(300×32×1)(1×16)=600 K (a)Work done in a polytropic processW=(P1V1−P2V2)(n−1)=R(T1−T2)(n−1) W=8.314(300−600)(1.25−1)=−9.977 kJ/mol (b)Heat transferred in this processQ=W{(γ−n)/(γ−1)} =−9.977(1.4−1.25)/(1.4−1) =−3.742 kJ/mol

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