A heat engine using a monatomic gas follows the cycle shown in the pV diagram to

Question

A heat engine using a monatomic gas follows the cycle shown in the pV diagram to the right. The gas starts out at point 1 with a volume of 233 cm3, a pressure of 147 kPa, and a temperature of 317 K. The gas is held at a constant volume while it is heated until its temperature reaches 395 K (point 2). The gas is then allowed to expand adiabatically until its pressure is again 147 kPa (point 3). The gas is maintained at this pressure while it is cooled back to its original temperature of 317 K (point 1 again).

For each stage of this process, calculate in joules the heat Qin transferred to the gas, and the work Wout done by the gas.

Explanation / Answer

Initially using PV = nRT

147 x 10^3 x 233 x 10^-6 = n x 8.314 x 317

n = 0.013 mol

for 1st process,

constant volume do Wout = 0 ......Ans

Qin = (3/2) nR deltaT = 1.5 x 8.314 x 0.013 x (395 - 317) = 12.64 J ....Ans

For 2nd Process,

Work done = P*V^y ( Vf^(1-y) - Vi^(1-y)) / 1-y

where y = Cp/Cv = 1.66 for monoatomic gas

and P*Vy = constant throughout the processs.

and Vf and Vi can be obtanied by

Vf = nRTf / Pf and Vi = nRTi / Pi

and Qin = 0 in adiabatic process

for 3 process,

pressure is maintained but temperature is changed so work done = P (deltaV) = nR(deltaT)

= 0.013 x 8.314 x (395 -317) =8.43 J

Qin = (5/2) nR(deltaT) = 8.43 x (5/2) = 21.08 J

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