Engineering and
Chemical
Thermodynamics 2nd
Edition By Milo D.
Koretsky (Solution
Manual)
2
,Chapter 1–2: Fundamentals of
Thermodynamics
Chapter 2: Energy Analysis of Closed
Systems
Sections 2.1–2.6 — Exam-Style Questions with Solutions
2.1 Energy Balance for a Closed System
Question 2.1
A closed system undergoes a process in which 80 kJ of heat is transferred to the system, and
30 kJ of work is done on the system.
Changes in kinetic and potential energy are negligible.
What is the change in internal energy?
A. −110 kJ
B. −50 kJ
C. +50 kJ
D. +110 kJ
Solution
First Law for a closed system:
ΔU=Q−W\Delta U = Q - WΔU=Q−W
Work done on the system → W=−30 kJW = -30 \, kJW=−30kJ
ΔU=80−(−30)=110 kJ\Delta U = 80 - (-30) = 110 \, kJΔU=80−(−30)=110kJ
Answer: D
2.2 Boundary Work
3
,Question 2.2
A gas expands in a piston–cylinder device at constant pressure of 200 kPa from 0.2 m³ to 0.6
m³.
What is the boundary work done by the gas?
A. 40 kJ
B. 60 kJ
C. 80 kJ
D. 120 kJ
Solution
Constant pressure work:
W=P(V2−V1)W = P(V_2 - V_1)W=P(V2−V1) W=200(0.6−0.2)=80 kJW = 200(0.6 - 0.2) = 80
\, kJW=200(0.6−0.2)=80kJ
Answer: C
2.3 Rigid Tank Heating
Question 2.3
A rigid tank contains a gas that is heated from 300 K to 600 K.
Which statement is TRUE?
A. Boundary work is positive
B. Boundary work is negative
C. Boundary work is zero
D. Boundary work depends on pressure
Solution
Rigid tank → volume constant → no boundary movement.
W=∫P dV=0W = \int P\, dV = 0W=∫PdV=0
Answer: C
4
, 2.4 Constant-Volume Process (Ideal Gas)
Question 2.4
A closed rigid tank contains 2 kg of air initially at 300 K. The air is heated to 900 K.
Given:
Cv=0.718 kJ\cdotpkg−1\cdotpK−1C_v = 0.718 \, \text{kJ·kg}^{-1}\text{·K}^{-1}Cv
=0.718kJ\cdotpkg−1\cdotpK−1
How much heat is transferred?
A. 430 kJ
B. 645 kJ
C. 860 kJ
D. 1,290 kJ
Solution
Rigid tank → Q=ΔUQ = \Delta UQ=ΔU
Q=mCv(T2−T1)Q = m C_v (T_2 - T_1)Q=mCv(T2−T1) Q=2×0.718×(900−300)Q = 2 \times
0.718 \times (900 - 300)Q=2×0.718×(900−300) Q=861.6 kJQ = 861.6 \, kJQ=861.6kJ
Answer: C
2.5 Constant-Pressure Process (Ideal Gas)
Question 2.5
A piston–cylinder contains 1 kg of an ideal gas at constant pressure.
The temperature increases from 300 K to 700 K.
Cp=1.005 kJ\cdotpkg−1\cdotpK−1C_p = 1.005 \, \text{kJ·kg}^{-1}\text{·K}^{-1}Cp
=1.005kJ\cdotpkg−1\cdotpK−1
How much heat is added?
A. 200 kJ
B. 302 kJ
5
Chemical
Thermodynamics 2nd
Edition By Milo D.
Koretsky (Solution
Manual)
2
,Chapter 1–2: Fundamentals of
Thermodynamics
Chapter 2: Energy Analysis of Closed
Systems
Sections 2.1–2.6 — Exam-Style Questions with Solutions
2.1 Energy Balance for a Closed System
Question 2.1
A closed system undergoes a process in which 80 kJ of heat is transferred to the system, and
30 kJ of work is done on the system.
Changes in kinetic and potential energy are negligible.
What is the change in internal energy?
A. −110 kJ
B. −50 kJ
C. +50 kJ
D. +110 kJ
Solution
First Law for a closed system:
ΔU=Q−W\Delta U = Q - WΔU=Q−W
Work done on the system → W=−30 kJW = -30 \, kJW=−30kJ
ΔU=80−(−30)=110 kJ\Delta U = 80 - (-30) = 110 \, kJΔU=80−(−30)=110kJ
Answer: D
2.2 Boundary Work
3
,Question 2.2
A gas expands in a piston–cylinder device at constant pressure of 200 kPa from 0.2 m³ to 0.6
m³.
What is the boundary work done by the gas?
A. 40 kJ
B. 60 kJ
C. 80 kJ
D. 120 kJ
Solution
Constant pressure work:
W=P(V2−V1)W = P(V_2 - V_1)W=P(V2−V1) W=200(0.6−0.2)=80 kJW = 200(0.6 - 0.2) = 80
\, kJW=200(0.6−0.2)=80kJ
Answer: C
2.3 Rigid Tank Heating
Question 2.3
A rigid tank contains a gas that is heated from 300 K to 600 K.
Which statement is TRUE?
A. Boundary work is positive
B. Boundary work is negative
C. Boundary work is zero
D. Boundary work depends on pressure
Solution
Rigid tank → volume constant → no boundary movement.
W=∫P dV=0W = \int P\, dV = 0W=∫PdV=0
Answer: C
4
, 2.4 Constant-Volume Process (Ideal Gas)
Question 2.4
A closed rigid tank contains 2 kg of air initially at 300 K. The air is heated to 900 K.
Given:
Cv=0.718 kJ\cdotpkg−1\cdotpK−1C_v = 0.718 \, \text{kJ·kg}^{-1}\text{·K}^{-1}Cv
=0.718kJ\cdotpkg−1\cdotpK−1
How much heat is transferred?
A. 430 kJ
B. 645 kJ
C. 860 kJ
D. 1,290 kJ
Solution
Rigid tank → Q=ΔUQ = \Delta UQ=ΔU
Q=mCv(T2−T1)Q = m C_v (T_2 - T_1)Q=mCv(T2−T1) Q=2×0.718×(900−300)Q = 2 \times
0.718 \times (900 - 300)Q=2×0.718×(900−300) Q=861.6 kJQ = 861.6 \, kJQ=861.6kJ
Answer: C
2.5 Constant-Pressure Process (Ideal Gas)
Question 2.5
A piston–cylinder contains 1 kg of an ideal gas at constant pressure.
The temperature increases from 300 K to 700 K.
Cp=1.005 kJ\cdotpkg−1\cdotpK−1C_p = 1.005 \, \text{kJ·kg}^{-1}\text{·K}^{-1}Cp
=1.005kJ\cdotpkg−1\cdotpK−1
How much heat is added?
A. 200 kJ
B. 302 kJ
5
