energy of the weight. Also by making necessary assumption of the geometry
of the piston cylinder determines the mass of the weight and the height
increase of the piston.
Q ¼ 518:4 kJ; DU ¼ 370:3 kJ
W atm ¼ 100 kJ; DPE weight ¼ 48:1 kJ
3:5 One gmol (i.e., mole) of a diatomic ideal gas performs a transformation from
an initial state for which temperature and volume are, respectively, 291 K
and 21 L to a final state in which temperature and volume are 305 K and
12.7 L. The transformation is represented on the (p, V) diagram by a straight
line*. Find the work performed and heat absorbed by the system.
W ¼ À1:3067 kJ; Q ¼ À1:0157 kJ
(work is done to the system and heat is rejected by the system)
3:6 Consider the same one gmol ideal gas at the same initial state of 291 K and
21 L, which is compressed to a final state volume of 12.7 L. What is the
initial pressure? What is the pressure and temperature at the final state if the
compression is adiabatic (i.e., isentropic—see Chap. 4) and the work performed? What is the pressure of the final state if the compression is
isothermal and the work performed and the heat absorbed by the system?
232:96kPa; 355:8K; À 1:347:9kJ; 190:5kPa; À1:2168kJ; À1:2168kJ
3:7 A diatomic ideal gas expands adiabatically* to a volume 1.35 times larger
than the initial volume. The initial temperature is 18 °C. Find the final
temperature.
258:2KðÀ14:9
CÞ
3:8 Two blocks A and B are initially at 100 and 500 °C, respectively. They are
brought together and isolated from the surroundings. Determine the final
equilibrium temperature of the blocks. Block A is aluminum
c p ¼ 0:900kJ=kg Á K
Â
Ã
with m A = 0.5 kg and block B is copper
c p ¼ 0:386kJ=kg Á K
Â
Ã
with m A = 1.0 kg. Note: for incompressible solids
(liquids too—see Sect. 9.5.2, Eq. (176B)), du % c p dT.
284:7
Cð557:85KÞ
3:9 In the adiabatic free expansion of a gas, how is it that the temperature doesn’t
drop, given that it increases when the gas is compressed back to its original
volume?
58
3 The First Law: The Production of Heat …
of the piston cylinder determines the mass of the weight and the height
increase of the piston.
Q ¼ 518:4 kJ; DU ¼ 370:3 kJ
W atm ¼ 100 kJ; DPE weight ¼ 48:1 kJ
3:5 One gmol (i.e., mole) of a diatomic ideal gas performs a transformation from
an initial state for which temperature and volume are, respectively, 291 K
and 21 L to a final state in which temperature and volume are 305 K and
12.7 L. The transformation is represented on the (p, V) diagram by a straight
line*. Find the work performed and heat absorbed by the system.
W ¼ À1:3067 kJ; Q ¼ À1:0157 kJ
(work is done to the system and heat is rejected by the system)
3:6 Consider the same one gmol ideal gas at the same initial state of 291 K and
21 L, which is compressed to a final state volume of 12.7 L. What is the
initial pressure? What is the pressure and temperature at the final state if the
compression is adiabatic (i.e., isentropic—see Chap. 4) and the work performed? What is the pressure of the final state if the compression is
isothermal and the work performed and the heat absorbed by the system?
232:96kPa; 355:8K; À 1:347:9kJ; 190:5kPa; À1:2168kJ; À1:2168kJ
3:7 A diatomic ideal gas expands adiabatically* to a volume 1.35 times larger
than the initial volume. The initial temperature is 18 °C. Find the final
temperature.
258:2KðÀ14:9
CÞ
3:8 Two blocks A and B are initially at 100 and 500 °C, respectively. They are
brought together and isolated from the surroundings. Determine the final
equilibrium temperature of the blocks. Block A is aluminum
c p ¼ 0:900kJ=kg Á K
Â
Ã
with m A = 0.5 kg and block B is copper
c p ¼ 0:386kJ=kg Á K
Â
Ã
with m A = 1.0 kg. Note: for incompressible solids
(liquids too—see Sect. 9.5.2, Eq. (176B)), du % c p dT.
284:7
Cð557:85KÞ
3:9 In the adiabatic free expansion of a gas, how is it that the temperature doesn’t
drop, given that it increases when the gas is compressed back to its original
volume?
58
3 The First Law: The Production of Heat …