A heat engine takes 0.350 mol of a diatomic ideal gas around the cycle shown in
ID: 1289872 • Letter: A
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.
Explanation / Answer
1.
p? and p? are given in the figure:
p? = p? = 1atm
To calculate p? use ideal gas law
pV = nRT
the the amount n stays throughout the whole process
pV/T = nR = 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 = nRT/p
V? = nRT?/p?
= 0.35mol 0.0820574587atmL/molK 300K / 1atm
= 8.616L
V? = V? = 8.616L
V? = nRT?/p?
= 0.35mol 0.0820574587atmL/molK 492K / 1atm
= 14.130L
3.
The change of internal energy of ideal gas is given by:
?U = nCv?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
pV^? = const
<=>
pV^? = 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
= - 2101325Pa 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