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This set of Engineering Physics Questions and Answers for Entrance exams focuses on “Isothermic and Adiabatic Process”.

1. Air in a cylinder is suddenly compressed by a piston which is then maintained at the same position. After sometime, the ___________
a) Pressure will increase
b) Pressure remains the same
c) Pressure will decrease
d) Pressure may increase or decrease

Explanation: Sudden compression increases inside temperature. After sometime heat flows out. This decreases temperature of the air, As V is constant, so pressure inside decreases.

2. Twp cylinders of equal size are filled with equal amount of ideal diatomic gas at room temperature. Both the cylinders are fitted with pistons. In cylinder A the piston is free to move, while in cylinder B the piston is fixed. When same amount of heat is added to cylinder A raises by 20 K. What will be the rise temperature of gas in cylinder B?
a) 28K
b) 20K
c) 15K
d) 10K

Explanation: For gas in cylinder A, Q = nCp ∆T1
For gas in cylinder B, Q = nCv ∆T2
∆T2=Cp/Cv ∆T1=7/5×20=28K.

3. A gas does work during adiabatic expansion. The source of mechanical energy so produced is the internal energy of the gas itself.
a) True
b) False

Explanation: By the first law of thermodynamics,
∆Q=∆U+∆W.
But for an isothermal process, ∆Q is zero, so ∆W=-∆U. Thus the energy required for doing mechanical work during adiabatic expansion is the internal energy of the gas itself.

4. Out of solid, liquid and gas of the same mass and at the same temperature, which one has the greatest internal energy?
a) Solid
b) Liquid
c) Gas
d) All three have the same internal energy

Explanation: Gas has the greatest internal energy because the potential energy (Which is negative) of the molecules is very small.

5. Out of solid, liquid and gas of the same mass and at the same temperature, which one has the least internal energy?
a) Solid
b) Liquid
c) Gas
d) All three have the same internal energy

Explanation: The one which has small potential energy has the greatest internal energy. On the other hand, the one which has large potential energy has the least internal energy; therefore the internal energy of the solid is least.

6. If Cp and Cv are the specific heats for a gas at constant pressure and at constant volume respectively, then the relation Cp-Cv=R is exact for?
a) Ideal and real gases at all pressures
b) Ideal gas at all pressures and real gas at a moderate pressure
c) Ideal gas and nearly true for real gases at high pressure
d) Ideal gas and nearly true for real gases at a moderate pressure

Explanation: Cp-Cv=R
Holds for ideal gas and for real gases at moderate pressure.

7. In an adiabatic process, the quantity which remains constant is ___________
a) Volume
b) Pressure
c) Temperature
d) Total heat of the system

Explanation: In an adiabatic process, no heat is exchanged between system and surroundings. Hence total heat of the system remains constant.

8. Assertion: In an isolated system, the entropy increases.
Reason: The processes in an isolated system are adiabatic.
a) Both assertion and reason are true and the reason is the correct explanation of the assertion
b) Both assertion and reason are true but the reason is not a correct explanation of the assertion
c) Assertion is true but the reason is false
d) Both assertion and reason are false

Explanation: Both the assertion and reason are true but the reason is not the correct explanation of the assertion. The entropy of an isolated system increases in accordance with the second law of thermodynamics.

9. The efficiency of a Carnot engine operating with reservoir temperatures kept at 100 °C and -23°C will be ___________
a) (100+23)/100
b) (100-23)/100
c) (373+250)/373
d) (373-250)/373

Explanation: Ƞ=(T1-T2)/T1
Ƞ=1-300/400=1/4=25%
Hence efficiency of 26% with real engine is impossible.

10. A Carnot engine, whose sink is at 300 K, has an efficiency of 40%. By how much should the temperature of source be increased, so as to increase the efficiency by 50% of original efficiency?
a) 380 K
b) 275 K
c) 325 K
d) 250 K

Explanation: Ƞ=1-T2/T1
0.4=1- 300/T1
T1=300/0.6=500 K
Increase in temperature of the source= (T1)-T1=750-500=250K.

Sanfoundry Global Education & Learning Series – Engineering Physics.

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