ExamVeda
Login
Home

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-10 section-5

Each section contains 100 MCQs on Chemical Engineering Thermodynamics .
91
The Carnot co-efficient of performance (COP) of a domestic air conditioner compared to a household refrigerator is
Discuss
Answer & Solution
Answer: Option A
Solution:
The COP of a domestic air conditioner is less than that of household refrigerator because the work input in case of domestic air conditioner is higher than refrigerator $${\bf{cop}} = \frac{{{\text{desired output}}}}{{{\text{work input}}}}.$$
92
The gas law (PV = RT) is true for an __________ change.
Discuss
Answer & Solution
Answer: Option C
Solution:
The gas law PV = RT is an ideal gas law and can be applied to an ideal gas that is undergoing any changes (isothermal, adiabatic etc.).
93
If the pressure on 100 c.c. of air is halved, then its volume (at the same temperature) would be __________ c.c.
Discuss
Answer & Solution
Answer: Option D
Solution:
Since, pressure and volume are inversely proportional on halving the pressure the volume will be doubled.
Hence volume = 200 c.c.
94
At 60° C, vapour pressure of methanol and water are 84.562 kPa and 19.953 kPa respectively. An aqueous solution of methanol at 60° C exerts a pressure of 39.223 kPa; the liquid phase and vapour phase mole fractions of methanol are 0.1686 and 0.5714 respectively. Activity co-efficient of methanol is
Discuss
Answer & Solution
Answer: Option A
Solution:
Given $${x_i} = 0.1686$$   and $${y_i} = 0.5714,\,P{\text{ }} = {\text{ }}39.233{\text{ }}Kpa$$       and $${p_{sat}} = 84.562\,Kpa$$
Since activity co-efficient of methanol $$ = \frac{{{y_i}P}}{{{x_i}p_i^{sat}}} = 0.152$$
95
Free energy change of mixing two liquid substances is a function of the
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\Delta {g^{mix}} = RT\sum\nolimits_{i = 0}^2 {{n_i}\ln {x_i}} $$
Hence, we can say free energy is dependent on concentration, temperature, quantities of constituents.
96
For an ideal solution, the value of activity co-efficient is
Discuss
Answer & Solution
Answer: Option B
Solution:
For ideal solution the value of activity co-efficient is 1.
Definition: activity co-efficient is defined as the ratio of fugacity at existing conditions to fugacity in ideal solution.
97
The freezing point of a liquid decreases when the pressure is increased, if the liquid __________ while freezing.
Discuss
Answer & Solution
Answer: Option A
Solution:
FROM the P-V-T diagram we can say that if the freezing point decreases with increase in pressure than liquid should contract during freezing.
98
A/an __________ system is exemplified by a vessel containing a volatile liquid in contact with its vapor.
Discuss
Answer & Solution
Answer: Option C
Solution:
Since the vessel containing liquid is transferring both mass and heat (energy) with its vapor we can consider it as open system.
99
During Joule-Thomson expansion of gases
Discuss
Answer & Solution
Answer: Option A
Solution:
During, joule-thomson effect basically the change in internal (inter-molecular) energy and change in pressure energy will be equal to zero hence, enthalpy will remain constant.
$$\eqalign{ & dU + d\left( {PV} \right) = 0 \cr & \Rightarrow d\left( {U + PV} \right) = 0 \cr & \Rightarrow d\left( H \right) = 0 \cr & {\text{or }}H = {\text{constant}}{\text{.}} \cr} $$
100
The entropy change in a reversible isothermal process, when an ideal gas expands to four times its initial volume is
Discuss
Answer & Solution
Answer: Option A
Solution:
For an isothermal process change in internal energy is equal to zero.
So, from first law
$$\eqalign{ & q = - w = - 2.303\,RT\,{\text{lo}}{{\text{g}}_{10}}\left( {\frac{{v2}}{{v1}}} \right) \cr & \Rightarrow q = - RT\,{\log _e}4 \cr & \Rightarrow {\text{We know for reversible isothermal process }}ds = - \frac{{\delta q}}{T} \cr & \Rightarrow {\text{So, }}ds = R\,{\log _e}4. \cr} $$