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Chemical Engineering Thermodynamics MCQ Questions & Answers | Chemical Engineering

Learn Chemical Engineering Thermodynamics MCQ questions & answers are available for a Chemical Engineering students to clear GATE exams, various technical interview, competitive examination, and another entrance exam. Chemical Engineering Thermodynamics MCQ question is the important chapter for a Chemical Engineering and GATE students. Page-3 section-1

Each section contains 100 MCQs on Chemical Engineering Thermodynamics .
21
In a P-V diagram (for an ideal gas), an isothermal curve will coincide withan adiabatic curve (through a point), when
Discuss
Answer & Solution
Answer: Option B
Solution:
The slope of isothermal process in $$p-v$$  plane is \[\frac{{dP}}{{dV}} = \frac{{ - P}}{V}\]
The slope of adiabatic process in $$P-V$$  plane is \[\frac{{dP}}{{dV}} = - \gamma \frac{P}{V}\]
\[ \Rightarrow \] Slopes become equal when \[\gamma = 1\]
22
The shape of T-S diagram for Carnot Cycle is a
Discuss
Answer & Solution
Answer: Option A
Solution:
The Carnot cycle consists of
$$1.$$ Reversible isothermal expansion so, \[T = {\rm{CONSTANT}}\]
$$2.$$ Reversible adiabatic expansion so, \[S\left( {{\rm{entropy}}} \right) = {\rm{CONSTANT}}\]
$$3.$$ Reversible isothermal compression so, \[T = {\rm{CONSTANT}}\]
$$4.$$ Reversible adiabatic compression so, \[S\left( {{\rm{entropy}}} \right) = {\rm{CONSTANT}}\]
23
With increase in compression ratio, the efficiency of the otto engine
Discuss
Answer & Solution
Answer: Option A
No explanation is given for this question. Let's Discuss on Board
24
What is the degree of freedom for a system comprising liquid water equilibrium with its vapour?
Discuss
Answer & Solution
Answer: Option B
Solution:
Given, vapor-liquid equilibrium :
Number of components \[\left( {{\rm{water}}} \right) = 1\]
Number of phases involved \[\left( {{\rm{vapor - liquid}}} \right) = 2\]
So, degree of freedom :
\[\begin{array}{l} \left( f \right) = C - \emptyset + 2\\ 1 - 2 + 2 = 1 \end{array}\]
25
__________ functions are exemplified by heat and work.
Discuss
Answer & Solution
Answer: Option A
Solution:
We know heat and work are path functions as they are dependent on path specified. But according to first law the difference between heat and work is an point function \[\delta Q - \delta W = dU\]
26
A two stage compressor is used to compress an ideal gas. The gas is cooled to the initial temperature after each stage. The intermediate pressure for the minimum total work requirement should be equal to the __________ mean of P1 and P2. (where, P1 and P2 are initial and final pressures respectively)
Discuss
Answer & Solution
Answer: Option C
No explanation is given for this question. Let's Discuss on Board
27
Helmholtz free energy (A) is defined as
Discuss
Answer & Solution
Answer: Option B
Solution:
Helmholtz free energy :
\[\begin{array}{l} A = U - TS\\ dA = - {W_R} \end{array}\]
So, the decrease in Helmholtz energy denotes maximum work done in an reversible process under isothermal conditions.
28
Gibbs free energy (F) is defined as
Discuss
Answer & Solution
Answer: Option B
Solution:
Gibbs free energy :
\[\begin{array}{l} G = H - TS\\ \Rightarrow - dG = {W_R} - PdV \end{array}\]
So, it represents maximum useful work in a reversible process.
29
A system is said to be at equilibrium, if the entropy of the system has reached __________ value.
Discuss
Answer & Solution
Answer: Option C
Solution:
According to second law of thermodynamics, the entropy \[{\left( {dS} \right)_{universe}} > 0,\]    for an irreversible or spontaneous process and \[d{S_{universe}} = 0\]   for a reversible process or almost equilibrium reached process. so for an natural process since there will be irreversibility always the system entropy goes on increasing until equilibrium is reached where the system entropy becomes equal to zero. Hence a maximum exists for a system.
30
In the reaction, C + O2 → CO2; ΔH = -94 kcal. What is the heat content (enthalpy) of O2?
Discuss
Answer & Solution
Answer: Option D
Solution:
The enthalpy of formation of an element in its pure state is zero .since it doesn’t require any energy to obtain this state ($$form$$) it is naturally occurring state of the element.
\[\begin{array}{l} {O_2}\left( g \right) \to {O_2}\left( g \right)\\ \Rightarrow \Delta H = 0 \end{array}\]