# Thermodynamics Mcqs

Q:

Total volume of a liquid vapour mixture is given by

 A) volume of the saturated liquid B) volume of the saturated vapour C) sum of volumes of saturated liquid and saturated vapour D) none of the mentioned

Answer & Explanation Answer: C) sum of volumes of saturated liquid and saturated vapour

Explanation: V=Vf+Vg.

3683
Q:

The Otto cycle consists of

 A) two reversible isotherms and two reversible isobars B) two reversible isochores and two reversible adiabatics C) two reversible isotherms and two reversible isochores D) two reversible isobars and two reversible adiabatics

Explanation: This can be shown in a p-V and T-s diagrams.

2713
Q:

For a process from state 1 to state 2, heat transfer in an irreversible process is given by

 A) Q for irreversible=(To)*(S1-S2) B) Q for irreversible>(To)*(S1-S2) C) Q for irreversible<(To)*(S1-S2) D) none of the mentioned

Explanation: To is the temperature of the surroundings and S1,S2 are the entropies at state 1 and 2 respectively ans ?S(universe)>0.

2285
Q:

The enthalpy of a substance(denoted by h), is defined as

 A) h=u-pv B) h=u+pv C) h=-u+pv D) h=-u-pv

Explanation: This is a basic definition for enthalpy.

2230
Q:

Specific volume of the mixture is given by

 A) (1+x)vf + (x)vg B) (1-x)vf + (x)vg C) (1-x)vf – (x)vg D) none of the mentioned

Explanation: Here vf=specific volume of saturated solid and vg=specific volume of saturated vapour.

2181
Q:

A piston-cylinder device initially contains air at 150 kPa and 27°C. At this state, the volume is 400 litre. The mass of the piston is such that a 350 kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine the total heat transferred to the air.

 A) 747 kJ B) 757 kJ C) 767 kJ D) 777 kJ

Explanation: Qin – Wout = ?U = m(u3 –u1)
m = P1V1/RT1 = 0.697 kg
u1 = u@ 300 K = 214.36 kJ/kg
u3 = u@ 1400 K = 1113.43 kJ/kg
Therefore Qin = 767 kJ.

2105
Q:

In a thermally insulated kitchen, a refrigerator with 2 kW motor for running the compressor provides 6000 kJ of cooling to the refrigerated space during 30 min operation. If the condenser coil placed behind the refrigerator rejects 8000 kJ of heat to the kitchen during the same period, calculate the change in internal energy of the kitchen.

 A) 3600 kJ B) 2400 kJ C) 4800 kJ D) none of the mentioned

Explanation: QKitchen = 0 (Insulated!),
W(Electrical) = – P*?? = – 2 kW*30*60 sec = – 3600 kJ
(It is negative because work is done on to the system)
Change in internal energy of the kitchen (?UKitchen) = QKitchen – WElectrical
= 0 – (–3600) = 3600 kJ.

1972
Q:

The value of constant of proportionality, J, has the value

 A) 1 B) 0 C) -1 D) infinity