A heat engine with 0.200 mol of a monatomic ideal gas initially fills a 3000 3 c
ID: 1585050 • Letter: A
Question
A heat engine with 0.200 mol of a monatomic ideal gas initially fills a 3000 3 cylinder Part A at 800 K. The gas goes through the following closed cycle Isothermal expansion to 6000 cm3 Isochoric cooling to 400 K sothermal compression to 3000 cm How much work does this engine do per cycle? Express your answer with the appropriate units Isochoric heating to 800 HK Value Units Submit uest Ans Part B What is its thermal efficiency? Express your answer with the appropriate units Submit Request AnswerExplanation / Answer
(1)
Work done BY the gas is given by the integral
W = p dV from initial to final volume
In the isochoric steps of the process no work is done, cause volume does no change, i.e
W = W = 0
For an ideal gas undergoing an isothermal change of state:
pV = nRT = constant
Hence:
pV = p_initial V_initial
<=>
p = p_initial V_initial/V
So the work done in thermal process is
W = p_initialV_initial 1/V dV from initial to final volume
= p_initialV_initial ln(V_final/V_initial)
Using ideal gas law, you can substitute p and V:
p_initialV_initial = nRT_initial
Hence:
W = nRT_initial ln(V_final/V_initial)
For the expansion process.
W = 0.2mol 8.3145J/molK 800K ln(6000cm³/3000cm³)
= 922.1 J
W2 = 0 J = W4
Work done by the gas in compression process is
W = 0.2mol 8.3145J/molK 400K ln(3000cm³/6000cm³)
= -461.1 J
So the net work done by the engine per cycle is:
W = W + W + W + W
= 461 J
2)
To calculate efficiency you need the heat transferred to the engine.
You find it from internal energy considerations
Change of internal energy of an ideal gas is given by:
U = nCvT
For a monatomic gas
Cv = (3/2)R
On the other hand change of internal energy equals heat transferred to the gas minus work done By it (or plus work done on it):
U = Q - W
For the isothermal steps
U = Q - W = 0
=>
Q = W
For the isochoric steps
W = 0
=>
Q = U = (3/2)nRT
Hence:
Q = W = 922.1 J
Q = (3/2)0.2mol8.3145J/mol (400K - 800K) = -997.74 J
Q = W = -461.1 J
Q = (3/2)0.2mol8.3145J/mol (800K - 400K) = +997.74 J
So the total heat transferred to the engine is
Q_in = Q+ Q = 1919.84 J
It efficiency is the ratio of net work done to heat absorbed:
= W / 1919.84
= 461 / 8454J
= 0.240
= 24.0%