Question

A heat engine takes 0.350 mol of a diatomic ideal gas around the cycle shown in the pV-diagram of the figure (Figure 1) . Process 1?2 is at constant volume, process 2?3 is adiabatic, and process 3?1 is at a constant pressure of 1.00 atm. The value of ? for this gas is 1.40.

Part B

Find the volume at points 1, 2, and 3.

Enter your answers numerically separated by commas.

Part C

Calculate Q, W , and ?U for the process 1?2.

Enter your answers numerically separated by commas.

Part D

Calculate Q, W , and ?U for the process 2?3.

Enter your answers numerically separated by commas.

Part E

Calculate Q, W , and ?U for the process 3?1.

Enter your answers numerically separated by commas.

Answer 1

1.
p? and p? are given in the figure:
p? = p? = 1atm

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? = 1atm � 600K/300K = 2atm


2.
use ideal gas law to calculate the volume:
V = n�R�T/p

V? = n�R�T?/p?
= 0.35mol � 0.0820574587atmL/molK � 300K / 1atm
= 8.616L

V? = V? = 8.616L

V? = n�R�T?/p?
= 0.35mol � 0.0820574587atmL/molK � 492K / 1atm
= 14.130L


3.
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??= n�[R/(?-1)]�(T? - T?)
= 0.35mol � [8.314472J/molK/(1.4-1) ] � (600K - 300K)
= 2182.5J

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

Therefore
?U?? = Q?? + W??
=>
Q?? = ?U?? - W?? = 2182.5J


4.
The work done one the gas is:
W?? = - ?V??V? p dV
for an adiabatic process
p�V^? = const
<=>
p�V^? = p?�V?^?
<=>
p = p?�V?^? �V^-?
Hence:
W?? = - ?V??V? p?�V?^? �V^-? dV
= - p?�V?^? � ?V??V? 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 Pam�=J
= - 2�101325Pa � 0.008616m� � (1/(1.4-1)) � [1 - (300K / 492K )^(1.4-1)]
= -783.7J

for adiabatic process:
Q?? = 0

Hence
?U?? = W?? = -783.7J


5.
?U??= n�[R/(?-1)]�(T? - T?)
= 0.35mol � [8.314472J/molK/(1.4-1) ] � (300K - 492K)
= -3990.9J

W?? = - ?V??V? p dV = 0
because p=constant = p?=p?
W?? = - p?�(V? - V?)
= p?�(V? - V?)
= 101325Pa � (0.014130m� - 0.008616m�)
= 558.7J

Q?? = ?U?? - W??
= -3990.9J - 558.7J
= -4549.6J