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 Multiple Choice QuestionsMultiple Choice Questions

21.

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

  • 1/2

  • 1/4

  • 1/3

  • 1/3

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22.

A system goes from A to B via two processes I and II as shown in the figure. If ∆U1 and ∆U2 are the changes in internal energies in the processes I and II respectively, the

  • ∆U1 = ∆U2

  • relation between ∆U1 and ∆U2 can not be determined

  • ∆U2 > ∆U1

  • ∆U2 > ∆U1

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23.

Which of the following statements is correct for any thermodynamic system?

  • The internal energy changes in all processes.

  • Internal energy and entropy are state functions.

  • The change in entropy can never be zero.

  • The change in entropy can never be zero.

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24.

Two thermally insulated vessels 1 and 2 are filled with air at temperatures (T1, T2), volume (V1, V2) and pressure (P1, P2) respectively. If the valve joining two vessels is opened, the temperature inside the vessel at equilibrium will be

  • T1 + T2

  • T1 + T2/2

  • fraction numerator straight T subscript 1 straight T subscript 2 space left parenthesis straight P subscript 1 straight V subscript 1 plus straight P subscript 2 straight V subscript 2 right parenthesis over denominator straight P subscript 1 straight V subscript 1 straight T subscript 2 space plus space straight P subscript 2 straight V subscript 2 straight T subscript 1 end fraction
  • fraction numerator straight T subscript 1 straight T subscript 2 space left parenthesis straight P subscript 1 straight V subscript 1 plus straight P subscript 2 straight V subscript 2 right parenthesis over denominator straight P subscript 1 straight V subscript 1 straight T subscript 2 space plus space straight P subscript 2 straight V subscript 2 straight T subscript 1 end fraction


C.

fraction numerator straight T subscript 1 straight T subscript 2 space left parenthesis straight P subscript 1 straight V subscript 1 plus straight P subscript 2 straight V subscript 2 right parenthesis over denominator straight P subscript 1 straight V subscript 1 straight T subscript 2 space plus space straight P subscript 2 straight V subscript 2 straight T subscript 1 end fraction

There will be no change in number of moles if the vessels are joined by the valve. Therefore, from gas equation

PV space equals space nRT
fraction numerator straight P subscript 1 straight V subscript 1 over denominator RT subscript 1 end fraction space plus space fraction numerator straight P subscript 2 straight V subscript 2 over denominator RT subscript 2 end fraction space equals space fraction numerator straight P left parenthesis straight V subscript 1 space plus straight V subscript 2 right parenthesis over denominator RT end fraction space
rightwards double arrow space fraction numerator straight P subscript 1 straight V subscript 1 straight T subscript 2 space plus space straight P subscript 2 space straight V subscript 2 straight T subscript 1 over denominator straight T subscript 1 straight T subscript 2 end fraction space equals fraction numerator space straight P left parenthesis straight V subscript 1 space plus straight V subscript 2 right parenthesis over denominator straight T end fraction
space straight T space equals space fraction numerator straight P left parenthesis straight V subscript 1 space plus straight V subscript 2 right parenthesis straight T subscript 1 straight T subscript 2 over denominator left parenthesis straight P subscript 1 straight V subscript 1 straight T subscript 2 space plus space straight P subscript 2 straight V subscript 2 straight T right parenthesis end fraction

Now, according to Boyle's law (pressure = constant) P1 V1 + P2 V2 = P(V1 + V2 )

Hence comma space straight T equals space fraction numerator left parenthesis straight P subscript 1 straight V subscript 1 space plus straight P subscript 2 straight V subscript 2 right parenthesis straight T subscript 1 straight T subscript 1 over denominator straight P subscript 1 straight V subscript 1 straight T subscript 1 space plus space straight P 2 straight V subscript 2 straight T subscript 1 end fraction

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25.

Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

  • (a) 195 K (b) 2.7 kJ

  • (a) 189 K (b) 2.7 kJ

  • (a) 195 K (b) –2.7 kJ

  • (a) 189 K (b) – 2.7 kJ


26.

Match the following

     

  • I-4, II-3, III-2, IV-1

  • I-3, II-2, III-1, IV-4

  • I-1, II-2, III-3, IV-4

  • I-4, II-2, III-3, IV-1


27.

The Zeroth law of thermodynamics leads to the concept of

  • internal energy

  • heat content

  • pressure

  • temperature


28.

The Carnot cycle of a reversible heat engine consists of

  • one isothermal and two adiabatic processes

  • two isothermal and one adiabatic processes

  • two isothermal and two adiabatic processes

  • two isobaric and two isothermal processes


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29.

A rigid container with thermally insulated walls contains a gas and a coil of resistance 50 Ω, carrying a current of 1 A. The change in internal energy of the gas after 2 min will be

  • 6 kJ

  • 10 kJ

  • 3 kJ

  • 12 kJ


30.

Two soap bubbles each with radius r1 and r2 coalesce in vacuum under isothermal conditions to form a bigger bubble of radius R. Then, R is equal to

  • r12 + r22

  • r12 - r22

  • r1 + r2

  • r12 + r222


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