Question

a.) In an Isothermal process the temperature stays thes same as the volume and pressure are allowed to change. In such a proces the work is found by W=nRTln (VfVi), with n as the number of moles, R as the constant 8.31 J/mole*K . How much work is done in an isothermal process of an ideal gas starting at a pressure of 2.10E2 kPa, and 0.0360 m3 volume as it expands to a volume of 0.165 m3?

b.) If the gas is then compressed isobarically back to its initial volume and finally returns to the initial pressure and volume ischorically. What is the total work done in this cycle?

Answer #1

a) By ideal gas equation, PV = nRT

nRT = PV = 2.10*(10^2)*(10^3)*0.0360 = 7560

Work done, w1= nRT ln(vf/vi) = 7560 ln (0.165/0.0360)= 11509.5 J

This work is done by the system

b) P1 V1 = P2 V2

Final pressure, P2= P1V1/ V2 = 7560/0.165 = 45.8 * (10^3) Pa

Work done during isobaric process, W2 = P ∆V = 45.8*(10^3)* (0.165-0.0360) = 5.91* (10^3) = 5910 J

This work is done on the system.

Work done during isochoric process is zero.

W1 = -11509.5 J & W2 = 5910 J ( Based on convention, work done on the system is positive & work done by the system is negative)

So net work done, W= w1+w2 = -11509.5+5910 = -5600 J

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