The working substance of an engine is 1.00 mol of a diatomic ideal gas. The engi
ID: 3892096 • Letter: T
Question
The working substance of an engine is 1.00 mol of a diatomic ideal 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 28.8 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.
Im not getting the right answer
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Explanation / Answer
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 ? (28.8/9)^1.4 = 5.096 atm = 516225 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) ? 101300 Pa ? (9
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