Answers and comments for the review activities and problems in Chapter 7.


2. 1.2 x 10-20J, 490m/s

4. 3.95 x 105Pa

6. It will remain the same.

8. 5.13m/s

10. n = 456

12. 224J

14. 4.0 liters

16. 1.02 x 10-8N

18. 28.3o


2. 2.7 x 1016

3. The 'total' internal energy would have to include the nuclear energy and the energy equivalent of the rest mass. These remain unchanged in 'normal', i.e. thermal processes, and they are therefore not included when we use the term 'internal energy'.

4. 4.9 liters

6. p/T remains constant, T changes from 273K to 373K. p goes to 1.37 pA

T is proportional to the average kinetic energy, i.e. to vrms2. As T and p go up, vrms2 goes up by a factor of 1.37 and vrms by a factor of 1.17.

(0.3)(1.5R)(100)J = 374J of energy is transferred from the water to the gas.

8. 877J


3. c

5. If the line were to go through the origin, p and T would be proportional and V = nRT/p would remain the same. If we assume that the origin of the graph is at absolute zero, p goes up by a larger factor than T, so that V decreases.

7. Both c and d are correct. During the time described by the first horizontal line the sample is partially solid and partially liquid. This is therefore true after 5 minutes.

The initial slope is proportional to the heat capacity of the solid. It is larger than the slope of the third segment, which is proportional to the heat capacity of the liquid.

9. b