A heat engine using a monatomic gas follows the cycle shown in the pV diagram to
ID: 1393718 • Letter: A
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 I with a volume of 318 cm^3, a pressure of 147 kPa, and a temperature of 317K. 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 Q in transferred to the gas, and the work W out done by the gas.Explanation / Answer
from iead gas euation
P*V = n*R*T
P1*V1 = n*R1*T1
n = P1*V1/R*T1 = (147000*318*10^-6)/(8.314*317) = 0.018 moles
step 1 ----> 2
volume is constant. i.e dV = 0
dw = P*dV = 0 <-------answer
P1 / P2 = T1 / T2
P2 = (T2*P1)/T1
P2 = (395*147)/317 = 183.17 kPa
v2= V1 = 318 cm^3
from 1st law of thermodynamics dQ = dU + dw = n*cv*dT + dw
dQ = n*cv*dT
cv = R/(gamma-1)
gamma = ratio of specific heat = 1.66
dQ = (0.018*8.314*(395-317))/(1.66-1) = 17.7 J <-------answer
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step 2--->3
in adiabatic proess no heat is supplied so Qin = 0
P3 = P1 = 147 kPa
P3*V3^gamma = P2*v2^gamma
(147*v3^gamma) = (183.17*318^1.66)
v3 = 363.1 cm^3
T3^gamma/P3^(gamma-1) = T2^gamma/P2^(gamma-1)
T3^1.66/(147^0.66) = 395^1.66/(183.17^0.66)
work done = (P2*v2 - P3*V3)/(gamma-1)
T3 = 361.2 K
work = ((183170*318*10^-6) - (147000*363.1*10^-6))/(1.66-1)
work = 7.38 J
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step 3 ----> 1
here pressure P3 = 147 kPa is conatant
dU = n*cp*dT = n*R*gamma*(T1-T3)/(gamma-1)
dU = (0.018*8.314*1.66*(361.9-317))/(1.66-1) = 16.9 J
dW = P3*dV = P3*(V3-V2)
dW = 147000*(363.1*10^-6-318*10^-6)
dW = 6.63 J <------------answer
dQ = dU + dw = 16.9 + 6.63 = 23.53 J <------------answer