A PV-diagram for a Carnot engine is shown. The working substance for this engine

Question

A PV-diagram for a Carnot engine is shown. The working substance for this engine can be a monatomic or diatomic ideal gas. Given that W and W' represent the work done by one mole of the monatomic and diatomic gas, respectively, calculate W'/W. (note: the correct answer is not 1)

A PV-diagram for a Carnot engine is shown. The working substance for this engine can be a monatomic or diatomic ideal gas. Given that W and W' represent the work done by one mole of the monatomic and diatomic gas, respectively, calculate W'/W. (note: the correct answer is not 1)

Explanation / Answer

Both gases undergo the same the change of temperature. You can show this by ideal gas law:
p?V = n?R?T
<=>
T = p?V/(n?R)
since only V changes
?T = p??V/(n?R)
You could calculate ?T from the given information, but you don't need it. Actually the solution is the same for any constant pressure process.

For constant pressure process change of enthalpy is equal to the heat transferred to the gas:
?H = Q

This change is different for the gases. Enthalpy of an ideal gas is solely a function of temperature:
H = n?Cp?T
Hence:
?H = n?Cp??T

The theoretical values for Cp
Cp? = 5/2?R (monatomic gas)
Cp? = 7/2?R (diatomic gas)

Therefore
Q?/Q? = ?H?/?H?
= (n?Cp???T) / (n?Cp???T)
= Cp?/Cp?
= 7/5


You can also show arguing with using internal energy and specific heat capacities at constant volume:
U = n?Cv?T
=>
?U = n?Cv??T
with the theoretical values
Cv? = 3/2?R (monatomic gas)
Cv? = 5/2?R (diatomic gas)

Moreover
?U = Q + W

The work done on an ideal gas in a constant pressure process is:
W = - p??V = - n?R??T
which is the same for both gases.

So
Q = ?U - W = n?Cv??T + - n?R??T = n?(Cv+R)??T
=>
Q?/Q? = (n?(Cv?+R)??T) / (n?(Cv?+R)??T)
= (Cv?+R) / (Cv?+R)
= 7/5

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