Question

The working substance of an engine is 1.00 mol of diatomic gas. The engine operates in a cycle consisting of three steps: (1) an adiabatic expansion from an initial volume of 9.00 L to a pressure of 1.00 atm and a volume of 30.6 L, (2) a compression at constant pressure to its original volume of 9.00 L, and (3) heating at constant volume to its original pressure. Find the efficiency of this cycle.

Answer 1

The thermal efficiency of a cycle process is ratio the net work done by the engine to the heat input.
? = W/Q_in

To solve this problem it is sufficient to consider the only the heat transfer in each step of the process.
For a cycle process the overall change in internal energy is zero:
?U = Q - W = 0
So net heat flow to and work done by the engine have same magnitude:
W = Q = Q_in - Q_out
Hence,
? = (Q_in - Q_out)/Q_in = 1 - (Q_out/Q_in)

The process consists of three steps:

- Step (1)
This step is adiabatic, so there is no heat transfer:
Q? = 0
For the following calculations you need to find the initial pressure of this process.
For a reversible, adiabatic process
P?V^? = constant
=>
P_initial?V_initial^? = P_final?V_final^?
=>
P_initial = P_final ? (V_final/V_initial)^?
? is the heat capacity ratio. For a diatomic gas.
? = Cp / Cv = 7/5 = 1.4
Hence,
P_initial = 1 atm ? (39.6/9)^1.4 = 7.96 atm = 806,400 Pa

- Step (2)
The heat transferred to a system in a constant pressure process equals its change in enthalpy. Change in enthalpy for an ideal gas is given by:
?H = n?Cp??T
Using ideal gas law you can express change in temperature in terms of change in volume:
T = P?V/(n?R)
=>
?T = (P/(n?R))??V
=>
?H = (Cp/R)?P??V
Molar heat capacity for a diatomic ideal gas is (7/2)?R. Therefore
Q? = ?H? = (7/2)?P??V
= (7/2) ? 1?101325 Pa ? (9