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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)(600-240) = 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 ~ -12To 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] = (18-12) / (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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