Assignment #6
Answers:
A) State B has the greatest internal
energy.
B) ΔE= ΔE1 + ΔE2
ΔE= ΔE3 + ΔE4
C) ΔE= ΔE1 – ΔE3
D) Above state A
5.7 You may have noticed that when you compress the air in a bicycle pump, the body of the
pump gets warmer. (a) Assuming the pump and the air in it comprise the system, what is the sign
of w when you compress the air? (b) What is the sign of q for this process? (c) Based on your
answers to parts (a) and (b), can you determine the sign of _E for compressing the air in the
pump? If not, what would you expect for the sign of ΔE? What is your reasoning?
A) The sign of w is (+).
B) The sign of q is (-).
C) Because the sign of w is positive and the sign of q is negative, we can't be sure what the
sign of ΔE is. Because the amount of heat lost is likely to be substantially less than the
amount of work done on the system, the sign of ΔE would be almost probably positive.
Answers:
A) Open system
B) Closed system
5.25 Calculate ΔE and determine whether the process is endothermic or exothermic for the
following cases: (a) q = 0.763 kJ and w = -840 J. (b) A system releases 66.1 kJ of heat to its
surroundings while the surroundings do 44.0 kJ of work on the system.
Answer:
a) ΔE= -0.077 kJ-Exothermic
Solution:
ΔE = q+ w
Since q= 0.763KJ and w=-840j= -0.84KJ
ΔE= -0.077KJ
The process is exothermic since it releases heat to the surrounding ΔE is negative.
b) E= -22.1 kJ Exothermic
Solution:
q= - 66.1KJ negative because its heat released to the surrounding
w= 44.0KJ work done on the system by the surrounding.
ΔE= q+w
ΔE= -66.1KJ + 44.0KJ
ΔE= -22.1KJ
The process is exothermic because the value of E is negative.
5.26 For the following processes, calculate the change in internal energy of the system and
determine whether the process is endothermic or exothermic: (a) A balloon is cooled by
removing 0.655 kJ of heat. It shrinks on cooling, and the atmosphere does 382 J of work on the
balloon. (b) A 100.0-g bar of gold is heated from 25 °C to 50 °C during which it absorbs 322 J of
heat. Assume the volume of the gold bar remains constant.
A) The cooling removes energy from the balloon .655kJ = -655 J
The work done on the balloon increases the energy of the balloon
ΔE= -655 J+ 382 J
ΔE= -273 J or -0.273 kJ
B) Q= mCΔT
322 J= (100 g) C (50°C – 25 °C)
C= 322 J/ (100 g)(25 °C)
C= 322 J/ 2500 g°C
C= 0.1288 J/ g°C