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

Question

A heat engine using a diatomic gas follows the cycle shown in the pV diagram to the night. the gas starts out at point 1 with a volume of 233 cm^3, a pressure of 235 kPa, and a temperature of 287 K. the gas is held at a constant volume while it is heated until its temperature reaches 455 K. the gas is then allowed to expand adiabatically until its pressure is again 235 kPa. the gas is maintained at this pressure while it is cooled back to its original temperature of 287 K 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

1) The change of internal energy of ideal gas is given by:
U = n·Cv·T

For an ideal gas
Cp - Cv = R
with = Cp/Cv
<=>
Cv = R/(-1)

U= [R/(-1)]·(T - T)
= [8.314472J/molK/(1.4-1) ] · (455K - 287K)
= 3491.88 J

The work done one the gas is:
W = - VV p dV = 0
because the volume does not change

Therefore
U = Q + W
=>
Q = U - W = 3491.88 J


2)p and p are given in the figure:
p = p = 235 kpa

To calculate p use ideal gas law
p·V = n·R·T
the the amount n stays throughout the whole process
p·V/T = n·R = constant
Hence
p·V/T = p·V/T
because V= V
p/T = p/T
=>
p = p·T/T = 235 kpa · 455K/287K = 372.56 kpa

T = Vp/n·R
= 0.000233*235 kpa / 0.0820574587atmL/molK
= 667 K

The work done one the gas is:
W = - VV p dV
for an adiabatic process
p·V^ = const
<=>
p·V^ = p·V^
<=>
p = p·V^ ·V^-
Hence:
W = - VV p·V^ ·V^- dV
= - p·V^ · VV V^- dV
= - p·V^ · (1/(1-)) · [ V^(1-) - V^(1-)]
= - p·V · (1/(-1)) · [1 - (V/V)^(-1)]
= - p·V · (1/(-1)) · [1 - (T/T)^(-1)]
here calculate in SI units because Pam³=J
= - 372.56 kpa · 0.000233m³ · (1/(1.4-1)) · [1 - (455K / 667K )^(1.4-1)]
= -30.78J

for adiabatic process:
Q = 0

Hence
U = W = -30.78J


3)
U= n·[R/(-1)]·(T - T)
= [8.314472J/molK/(1.4-1) ] · (287K - 667K)
= -7898.68J

W = - VV p dV = 0
because p=constant = p=p
W = - p·(V - V)
= p·(V - V)
= 235 kPa · (0)
= 0 J

Q = U - W
= -7898.68J

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