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Homework Statement
A heat engine takes 4.0 moles of an ideal gas through the reversible cycle abca, on the pV diagram, as shown. The path bc is an isothermal process. The temperature at c is 600 K and the volumes at a and c are 0.04 m^3 and 0.10m^3, respectively. The molar heat capacity at constant volume, of the gas, is 30 J/mol*K. In the figure, the thermal efficiency of the engine is closest to:
a) 0.07
b) 0.10
c) 0.15
d) 0.22
e) 0.30
Homework Equations
[tex]P = nRT[/tex]
[tex]Q = nC_{v}\Delta[/tex]T
[tex]Q = nC_{p}\Delta[/tex]T
W = P[tex]\Delta[/tex]V
[tex]W = nRT ln \frac{V_{f}}{V_{i}}[/tex]
[tex]e = \frac{W}{Q_{h}} = 1  \frac{Q_{c}}{Q_{h}}[/tex]
The Attempt at a Solution
First, I calculated the temperature and pressures at points a, b, and c using PV = nRT.
Point A:
V = 0.04 m^3
T = 600 K
P = (4)(8.31)(600) / 0.04 = 498000 Pa
Point B:
V = 0.04 m^3
T = PV / NR = (199440)(0.04) / (4)(8.31) = 240 K
P = 199440 Pa
Point C:
V = 0.10 m^3
T = 600 K
P = nRT / v = (4)(8.31)(600) / .10 = 199440
Then I calculated the heat absorbed on the path ab in kJ:
[tex]Q = nC_{v}\Delta[/tex]T
Q = (4)(30)(600240) = 43200 ~ 43
Then I calculated the heat absorbed on the path ca in kJ:
[tex]Q = nC_{p}\Delta[/tex]T
Q = (4)(30 + 8.31)(360) = 55166.4 ~ 55
Then I calculated the work done on the path bc in kJ:
[tex]W = nRT ln \frac{V_{f}}{V_{i}}[/tex]
(4)(8.31)(600)ln (.10/.04) = 18274.5 ~ 18
[/tex]
Then I calculated the work done on the path ca in kJ:
W = P[tex]\Delta[/tex]V
(199440)(.04.10) = 11966.4 ~ 12
To calculate the thermal efficiency, I tried this:
[tex]e = \frac{W}{Q_{h}} = 1  \frac{Q_{c}}{Q_{h}}[/tex]
[tex]e = \frac{W}{Q_{h}}[/tex] = (1812) / (43 55) = 0.5
Obviously that cannot be correct.
The correct answer is B. 0.10.
How should I calculate the efficiency?
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