Engineering Thermodynamics Exam Prep 2026 | Practice Questions & Complete Solutions

THERMODYNAMICS
Chapter 2: Properties of Pure Substances
Phase Change, Steam Tables, Quality & Ideal Gas
10 Questions | Easy · Medium · Hard | Full Worked Solutions


Questions at a Glance
Q# Topic Level Key
Concept
1 Phase Diagram Easy Phase

2 Quality of Wet Steam Medium Quality

3 Steam Tables — Superheated Medium Steam

4 Compressed Liquid Approximation Medium Compressed

5 Ideal Gas Law Easy Ideal

6 Dryness Fraction from First Principles Hard Dryness

7 Compressibility Factor Z Medium Compressibility

8 Enthalpy of Vaporization Medium Enthalpy

9 Triple Point Easy Triple

10 Van der Waals Equation Hard Van




Q1. Phase Diagram
[Easy]

Describe the three regions on a P-v diagram for a pure substance. What are the saturated liquid
and saturated vapor lines? What is the critical point?
✔ SOLUTION
THREE REGIONS on P-v diagram:
1. COMPRESSED (SUBCOOLED) LIQUID: Left of saturated liquid line.
High pressure relative to saturation; liquid phase.
2. SATURATION (TWO-PHASE) REGION: Under the dome (between sat. liquid & sat. vapor lines).
Liquid and vapor coexist; P and T are NOT independent (T_sat depends on
P).
3. SUPERHEATED VAPOR: Right of saturated vapor line.
Vapor above saturation temperature at given pressure.

SATURATED LIQUID LINE (x=0): Connects states where the last drop of vapor
condenses.
SATURATED VAPOR LINE (x=1): Connects states where the first drop of liquid
forms.

CRITICAL POINT: Where the two saturation lines meet. For water:
P_cr = 22.06 MPa, T_cr = 373.95°C
Above the critical point, no phase distinction exists.



Thermodynamics — Practice Q&A | Page 1 of 6

, KEY ANSWER: Compressed liquid | Two-phase dome | Superheated vapor |
Critical point: 22.06 MPa, 374°C


Q2. Quality of Wet Steam
[Medium]

Steam at 200 kPa has a quality of x = 0.75. Using steam tables: T_sat = 120.21°C, h_f = 504.7
kJ/kg, h_fg = 2201.6 kJ/kg, v_f = 0.001061 m³/kg, v_g = 0.8858 m³/kg. Find: (a) specific enthalpy,
(b) specific volume.
✔ SOLUTION
Quality x = 0.75 means 75% vapor, 25% liquid by mass.

Part (a) – Specific enthalpy:
h = h_f + x × h_fg
h = 504.7 + 0.75 × 2201.6
h = 504.7 + 1651.2
h = 2155.9 kJ/kg

Part (b) – Specific volume:
v = v_f + x × v_fg
v_fg = v_g – v_f = 0.8858 – 0.001061 = 0.8847 m³/kg
v = 0.001061 + 0.75 × 0.8847
v = 0.001061 + 0.6635
v = 0.6646 m³/kg
KEY ANSWER: h = 2155.9 kJ/kg | v = 0.6646 m³/kg


Q3. Steam Tables — Superheated
[Medium]

Steam is at 1.4 MPa and 350°C. Using steam tables (interpolation): at 1.4 MPa, T_sat ≈ 195°C. At 1
MPa/350°C: h = 3157.7 kJ/kg; at 2 MPa/350°C: h = 3137.0 kJ/kg. Find the enthalpy by interpolation.
✔ SOLUTION
Since T = 350°C > T_sat = 195°C at 1.4 MPa, the steam is SUPERHEATED.

LINEAR INTERPOLATION between P = 1 MPa and P = 2 MPa at T = 350°C:

h at 1.4 MPa = h(1 MPa) + [(1.4 – 1.0)/(2.0 – 1.0)] × [h(2 MPa) – h(1
MPa)]
h = 3157.7 + [(0.4)/(1.0)] × (3137.0 – 3157.7)
h = 3157.7 + 0.4 × (–20.7)
h = 3157.7 – 8.28
h ≈ 3149.4 kJ/kg

NOTE: Enthalpy of superheated steam decreases slightly with increasing pressure
at constant temperature, which is why h(2 MPa) < h(1 MPa) here.
KEY ANSWER: h ≈ 3149.4 kJ/kg (superheated steam at 1.4 MPa, 350°C)


Q4. Compressed Liquid Approximation

Thermodynamics — Practice Q&A | Page 2 of 6