Question #137272

A piston cylinder contains gas initially at 3500 kPa with a volume of 0.03 cubic meter. The gas
is compressed during a process where pV raise to 1.25 = C to a pressure of 8500 kPa. The heat
transfer from the gas is 2.5 kJ. neglecting changes in kinetic and potential energies determine the nonflow work in kJ.

Expert's answer

The polytropic process is one in which the pressure-volume relation is given as pVn=constantpV^n=constant

pV1.25=constant    p1V11.25=p2V21.25pV^{1.25} =constant\implies p_1V_1^{1.25}=p_2V_2^{1.25}

V2=V1(p1p2)11.25=0.03(35008500)11.25=0.01475m3V_2=V_1(\frac{p_1}{p_2})^{\frac{1}{1.25}}=0.03(\frac{3500}{8500})^{\frac{1}{1.25}}=0.01475 m^3

The work done during the polytropic process is found by substituting the pressure volume

relation into the boundary work equation. The result is W=12pdV=p2V2p1V11nW= \int_1^2pdV=\frac{p_2V_2-p_1V_1}{1-n}

W=8500×103×0.014753500×103×0.0311.25=81500JW=\frac{8500×10^3×0.01475-3500×10^3×0.03}{1-1.25}=-81500 J

Then, according to the first law of thermodynamics, ΔU=QW=2500+81500=79000J=79kJΔU=Q-W=-2500+81500=79000 J=79kJ


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