Chemical Engineering Thermodynamics MCQ Questions & Answers

11

Efficiency of a heat engine working on Carnot cycle between two temperature levels depends upon the

A) Two temperatures only B) Pressure of working fluid C) Mass of the working fluid D) Mass and pressure both of the working fluid

Discuss

Answer & Solution

Answer: Option A

Solution:

Efficiency of a Carnot engine is given as $$\left( {{T_1} > {T_2}} \right)\aleph = 1 - \frac{{{T_2}}}{{{T_1}}}$$
So, it's solely dependent on temperatures of the reservoirs.

12

Efficiency of a Carnot engine working between temperatures T₁ and T₂ (T₁ < T₂) is

A) $$\frac{{{{\text{T}}_2} - {{\text{T}}_1}}}{{{{\text{T}}_2}}}$$ B) $$\frac{{{{\text{T}}_2} - {{\text{T}}_1}}}{{{{\text{T}}_1}}}$$ C) $$\frac{{{{\text{T}}_1} - {{\text{T}}_2}}}{{{{\text{T}}_2}}}$$ D) $$\frac{{{{\text{T}}_1} - {{\text{T}}_2}}}{{{{\text{T}}_1}}}$$

Discuss

Answer & Solution

Answer: Option A

Solution:

Here the question should be read carefully given $$\left( {{T_1} < {T_2}} \right)$$
So, the hot reservoir is $${{T_2}}$$ and cold reservoir is $${{T_1}}$$ hence the efficiency will be $$\aleph = 1 - \frac{{{T_1}}}{{{T_2}}}$$

13

Co-efficient of performance for a reversed Carnot cycle working between temperatures T₁ and T₂ (T₁ > T₂) is

A) $$\frac{{{{\text{T}}_2}}}{{{{\text{T}}_1} - {{\text{T}}_2}}}$$ B) $$\frac{{{{\text{T}}_1}}}{{{{\text{T}}_1} - {{\text{T}}_2}}}$$ C) $$\frac{{{{\text{T}}_1} - {{\text{T}}_2}}}{{{{\text{T}}_1}}}$$ D) $$\frac{{{{\text{T}}_1} - {{\text{T}}_2}}}{{{{\text{T}}_2}}}$$

Discuss

Answer & Solution

Answer: Option A

Solution:

The reverse of heat engine is nothing but refrigerator coefficient of performance is given as $$COP = \frac{{{T_2}}}{{{T_1} - {T_2}}}$$

14

The following heat engine produces power of 100000 kW. The heat engine operates between 800 K and 300 K. It has a thermal efficiency equal to 50% of that of the Carnot engine for the same temperature. The rate at which heat is absorbed from the hot reservoir is

A) 100, 000 kW B) 160, 000 kW C) 200, 000 kW D) 320, 000 kW

Discuss

Answer & Solution

Answer: Option D

No explanation is given for this question. Let's Discuss on Board

15

Degress of freedom at triple point will be

A) 0 B) 1 C) 2 D) 3

Discuss

Answer & Solution

Answer: Option A

Solution:

For a triple point number of components $$= 1$$
Number of phases in equilibrium $$= 3$$
So, by Gibbs phase rule $$f = c - \emptyset + 2 = 1 - 3 + 2 = 0$$

16

Refrigeration cycle

A) Violates second law of thermodynamics B) Involves transfer of heat from low temperature to high temperature C) Both A and B D) Neither A nor B

Discuss

Answer & Solution

Answer: Option B

Solution:

Refrigeration is just the reverse of heat engine so, it takes heat from low temperature by taking the help of surroundings and transfers it to high temperature reservoir.

17

Number of components (C), phase (P) and degrees of freedom (F) are related by Gibbs phase rule as

A) P + F - C = 2 B) C = P - F + 2 C) F = C - P - 2 D) P = F - C - 2

Discuss

Answer & Solution

Answer: Option A

Solution:

The Gibbs phase rule for non-reacting system and for equilibrium system is given as $$F=C-P+2$$

18

High pressure steam is expanded adiabati-cally and reversibly through a well insulated turbine, which produces some shaft work. If the enthalpy change and entropy change across the turbine are represented by ΔH and ΔS respectively for this process:

A) ΔH = 0 and ΔS = 0 B) ΔH ≠ 0 and ΔS = 0 C) ΔH ≠ 0 and ΔS ≠ 0 D) ΔH = 0 and ΔS ≠ 0

Discuss

Answer & Solution

Answer: Option B

Solution:

For a reversible adiabatic process the entropy change is zero since $$dS = \frac{{\delta Q}}{T}$$
But the enthalpy change need not be zero since the internal energy and pressure energy may vary during the process.

19

Compressibility factor of a gas is

A) Not a function of its pressure B) Not a function of its nature C) Not a function of its temperature D) Unity, if it follows PV = nRT

Discuss

Answer & Solution

Answer: Option D

Solution:

The compressibility factor (Z) is defined as Z = PV / nRT, which measures the deviation of a real gas from ideal gas behavior.

For an ideal gas, the relation PV = nRT holds exactly. Substituting into the expression for Z gives Z = 1. Thus, the compressibility factor is exactly unity when the gas follows the ideal gas law.

Real gases, however, deviate from ideal behavior due to intermolecular forces and finite molecular volumes. In such cases, Z ≠ 1, and its value depends on the pressure, temperature, and nature of the gas.

Option A: Incorrect, because Z does vary with pressure for real gases.

Option B: Incorrect, because Z depends on the nature of the gas.

Option C: Incorrect, because Z also varies with temperature.

Option D: Correct, because Z = 1 only when the gas behaves ideally according to PV = nRT.

Therefore, the compressibility factor of a gas is unity if it follows PV = nRT.

20

In the equation, PVⁿ = Constant, if the value of n = 0, then it represents a reversible __________ process.

A) Isobaric B) Isothermal C) Isentropic D) Isometric

Discuss

Answer & Solution

Answer: Option A

Solution:

Pressure constant process is known as isobaric process.

Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering