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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 413: Standardized Test Prep

Exercise 1

Step 1

1 of 2

The gas compresses during process I so then is work done on the gas.

Result

2 of 2

(A)

Exercise 2

Step 1

1 of 2

The heat exchanged with the environment is equal to the change in thermal energy in steps II and IV because they are comprised of isochoric processes. Since we search for step where the thermal energy increases choice reduces to step II (the pressure is increasing at constant volume).

Result

2 of 2

(B)

Exercise 3

Step 1

1 of 2

Total work done is equal to the area of the square made by contours of the cycle:

$$
W=30text{ kPa}times 0.3text{ m}^3 =9000text{ J}.
$$

Result

2 of 2

(B)

Exercise 4

Step 1

1 of 2

We easily recognize its’ statement that says that the change in thermal energy of the system is equal to the heat exchanged with the environment and the work done by the system.

Result

2 of 2

(D)

Exercise 5

Step 1

1 of 2

There is no change in volume along the vertical line and thus no work done.

Result

2 of 2

(D)

Exercise 6

Step 1

1 of 2

The efficiency, by definition is the ratio of work done and heat added so it will increase in case (A)

Result

2 of 2

(A)

Exercise 7

Step 1

1 of 2

In this process the thermal energy must remain constant which is the case in isothermal process.

Result

2 of 2

(B)

Exercise 8

Step 1

1 of 2

In isothermal process there is no change in thermal energy so we easily conclude that the system does $350text{ J}$ of work.

Result

2 of 2

(C)

Exercise 9

Step 1

1 of 2

a) Increasing $T_h$ decreases the term $T_c/T_h$ which then increases the multiplier $1-T_c/T_h$ and thus the work.

b) Decreasing $T_c$ decreases $T_c/T_h$ thus increasing the multiplier $1-T_c/T_h$ and the work done.

c) It is directly seen that the work increases since $Q_h$ increases.

d) Because then the term $T_c/T_h$ would be zero and that requires $T_c = 0$ which is forbidden by the third law of thermodynamics.

Result

2 of 2

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