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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-9 section-2

Each section contains 100 MCQs on Chemical Engineering Thermodynamics .
81
Pick out the wrong statement pertaining to the decomposition of PCl5 represented by, PCl5 ⇋ PCl3 + Cl2. Degree of dissociation of PCl5 will
Discuss
Answer & Solution
Answer: Option D
Solution:
By Le Chatelier's principle we can say that the dissociation of $$PC{l_5}$$  is increased by decrease of number of moles of the products or by decreasing the pressure.
82
A gas mixture of three components is brought in contact with a dispersion of an organic phase in water. The degree of freedom of the system are
Discuss
Answer & Solution
Answer: Option B
No explanation is given for this question. Let's Discuss on Board
83
The compressibility factor of a gas is given by (where, V1 = actual volume of the gas V2 = gas volume predicted by ideal gas law)
Discuss
Answer & Solution
Answer: Option A
Solution:
The compressibility factor is given by $$Z = \frac{{P{V_1}}}{{RT}}.$$   If at the same temperature and pressure if the same gas exists as ideal gas than we can write:
\[\begin{array}{l} Z = 1 = \frac{{P{V_2}}}{{RT}} \Rightarrow \frac{{RT}}{P} = {V_2}\\ {\rm{So, }}\,\,Z = \frac{{{V_1}}}{{{V_2}}}. \end{array}\]
84
Air enters an adiabatic compressor at 300 K. The exit temperature for a compression ratio of 3, assuming air to be an ideal gas $$\left( {\gamma = \frac{{{{\text{C}}_{\text{p}}}}}{{{{\text{C}}_{\text{v}}}}} = \frac{7}{5}} \right)$$   and the process to be reversible, is
Discuss
Answer & Solution
Answer: Option A
Solution:
We know for an ideal gas at adiabatic conditions
\[\begin{array}{l} T{V^{\gamma - 1}} = {\rm{CONSTANT}}\\ \Rightarrow T{\left( {\frac{{RT}}{P}} \right)^{\gamma - 1}} = {\rm{CONSTANT}}\\ \Rightarrow T{P^{\frac{{1 - \gamma }}{\gamma }}} = {\rm{CONSTANT}}\\ {\rm{So, }}\,\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\gamma - 1}}{\gamma }}} \end{array}\]
$$ \Rightarrow $$  given compression ratio $$\frac{{{P_2}}}{{{P_1}}} = 3{\text{ and }}\gamma = \frac{7}{5}$$
$${\text{So, }}{T_2} = 300{\left( 3 \right)^{\frac{2}{7}}}.$$
85
As the temperature is lowered towards the absolute zero, the value of the quantity $$\left( {\frac{{\partial \Delta {\text{F}}}}{{\partial {\text{T}}}}} \right)$$  approaches
Discuss
Answer & Solution
Answer: Option A
No explanation is given for this question. Let's Discuss on Board
86
Refrigeration capacity of a household refrigerator may be round about __________ tons.
Discuss
Answer & Solution
Answer: Option A
No explanation is given for this question. Let's Discuss on Board
87
All gases during throttling process at atmospheric temperature and pressure show a cooling effect except
Discuss
Answer & Solution
Answer: Option B
Solution:
For almost all the gases the maximum inversion temperature is above the room temperature but for gases like $${H_2}.$$ The inversion temperatures are 200K and 24K respectively so by throttling always heating occurs to accomplish cooling they are brought below the inversion temperature by some medium like liquid nitrogen etc. Hence to cool the $${H_2}$$ simply throttling wont suffise.
88
On opening the door of an operating refrigerator kept in a closed room, the temperature of the room will
Discuss
Answer & Solution
Answer: Option A
Solution:
On keeping the door of a refrigerator open the refrigerator will give heat to the room and therefore the room temperature increases as the refrigerator takes heat from low temperature reservoir and gives it to high temperature reservoir here the room is closed so the refrigerator gives heat to the surrounding air in the room making the average temperature of the room increase.
89
The value of gas constant 'R' is
Discuss
Answer & Solution
Answer: Option C
Solution:
The given is value of universal gas constant in different units so, both are correct.
90
With increase in pressure (above atmospheric pressure), the Cp of a gas
Discuss
Answer & Solution
Answer: Option A
Solution:
The $${C_P}$$ increases because in case of higher pressure it has to do more work on the surroundings there by for on given amount of heat more amount of heat will be used to do work when compared to previous case where the pressure is slightly lower and hence more amount of heat has to be given to the system to raise the system to same temperature as of previous case, so the value of $${C_P}$$ increases.