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 Multiple Choice QuestionsMultiple Choice Questions

1.

Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume straight u equals space straight U over straight V space proportional to space straight T to the power of 4 and pressure.straight p space equals space 1 third open parentheses straight U over straight V close parentheses If the shell now undergoes an adiabatic expansion the relation between T and R is

  • T ∝ e-R

  • T ∝ e-3R

  • T ∝ (1/R)

  • T ∝(1/R3)


C.

T ∝ (1/R)

According to given equation,

straight p space equals space 1 third open parentheses straight U over straight V close parentheses
rightwards double arrow nRT over straight V space equals space 1 third open parentheses straight U over straight V close parentheses
left square bracket because space pV space equals space nRT right square bracket
or space nRT over straight V space proportional to 1 third straight T to the power of 4
or space VT cubed space equals constant
or space 4 over 3 πR cubed straight T cubed space equals space constant
or space TR space equals constant
straight T proportional to 1 over straight R

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2.

Helium gas goes through a cycle ABCDA (consisting of two isochoric and two isobaric lines) as shown in the figure. Efficiency of this cycle is nearly:(Assume the gas to be close to ideal gas)

  • 15.4%

  • 9.1%

  • 10.5%

  • 12.5%


A.

15.4%

The efficiency of a process is defined as the ratio of work done to energy supplied.
Here,straight eta space equals space fraction numerator increment straight W over denominator increment straight Q end fraction space equals space fraction numerator Area space under space straight p minus straight V space diagram over denominator increment straight Q subscript AB plus space increment straight Q subscript BC end fraction
Where Cp and Cv are two heat capacities (molar)
therefore space space fraction numerator straight p subscript straight o straight V subscript straight o over denominator nC subscript straight v increment straight T subscript 1 space plus nC subscript straight p space increment straight T subscript 2 end fraction
space equals fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 end style nR left parenthesis straight T subscript straight B minus straight T subscript straight A right parenthesis plus begin display style 5 over 2 end style nR left parenthesis straight T subscript straight C minus straight T subscript straight D right parenthesis end fraction
equals space fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 left parenthesis 2 straight p subscript straight o straight V subscript straight o minus straight p subscript straight o straight V subscript straight o right parenthesis plus 5 over 4 left parenthesis 4 straight p subscript straight o straight V subscript straight o minus 2 straight p subscript straight o straight V subscript straight o right parenthesis end style end fraction
equals space fraction numerator straight p subscript straight o straight V subscript straight o over denominator begin display style 3 over 2 straight p subscript straight o straight V subscript straight o plus 5 over 4 2 straight p subscript straight o straight V subscript straight o end style end fraction
equals space fraction numerator 1 over denominator 6.5 end fraction space equals space 15.4 percent sign

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3.

One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in the figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement:

  • The change in internal energy in the process AB is -350 R.

  • The change in internal energy in the process BC is -500 R.

  • The change in internal energy in the whole cyclic process is 250 R.

  • The change in internal energy in the process CA is 700 R.


D.

The change in internal energy in the process CA is 700 R.

According to first law of thermodynamics, 
(i) change in internal energy from A to B i.e, 

increment straight U subscript AB space equals space nC subscript straight V left parenthesis straight T subscript straight B minus straight T subscript straight A right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 800 minus 400 right parenthesis space equals space 1000 straight R
increment straight U subscript BC space equals space nC subscript straight V left parenthesis straight T subscript straight C minus straight T subscript straight B right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 600 minus 400 right parenthesis space equals space minus 500 straight R
increment straight U subscript Total space equals space 0
increment straight U subscript CA space equals space nC subscript straight V left parenthesis straight T subscript straight A minus straight T subscript straight C right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 400 minus 600 right parenthesis space equals space minus 500 straight R space

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4.

A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways:
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies the same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies the same amount of heat. In both the cases body is brought from an initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is:

  • ln2,4ln2

  • ln2,ln2

  • ln2,2ln2

  • 2ln2,8ln2


B.

ln2,ln2

Since entropy is a state function, therefore a change in entropy in both the processes must be same .

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5.

A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be

  • the efficiency of Carnot engine cannot be made larger than 50%

  • 1200 K

  • 750 K 

  • 600 K


C.

750 K 

Efficiency
straight eta space equals space 1 minus straight T subscript sink over straight T subscript source
Now comma space 0.4 space equals 1 minus fraction numerator straight T subscript sink over denominator 500 straight K end fraction
rightwards double arrow space straight T subscript sink space equals space 0.6 space straight x space 500 space straight K
space equals space 300 space straight K
Thus comma space 0.6 space equals space 1 minus fraction numerator 300 space straight K over denominator straight T apostrophe subscript source end fraction
rightwards double arrow space straight T to the power of apostrophe subscript source space equals space fraction numerator 300 space straight K over denominator 0.4 end fraction
space equals 750 space straight K

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6.

Three perfect gases at absolute temperature T1, T2 and T3 are mixed. The masses of molecules are m1,m2 and m3 and the number of molecules is n1,n2 and n3 respectively.Assuming no loss of energy, the final temperature of the mixture is

  • fraction numerator straight n subscript 1 straight T subscript 1 plus straight n subscript 2 straight T subscript 2 space plus straight n subscript 3 straight T subscript 3 over denominator straight n subscript 1 plus space straight n subscript 2 plus space straight n subscript 3 end fraction
  • fraction numerator straight n subscript 1 straight T subscript 1 superscript 2 space plus straight n subscript 2 straight T subscript 2 superscript 2 space plus straight n subscript 3 straight T subscript 3 superscript 2 over denominator straight n subscript 1 straight T subscript 1 space plus straight n subscript 2 straight T subscript 2 space plus straight n subscript 3 straight T subscript 3 end fraction
  • fraction numerator straight n squared subscript 1 straight T subscript 1 superscript 2 space plus straight n squared subscript 2 straight T subscript 2 superscript 2 space plus straight n squared subscript 3 straight T subscript 3 superscript 2 over denominator straight n subscript 1 straight T subscript 1 space plus straight n subscript 2 straight T subscript 2 space plus straight n subscript 3 straight T subscript 3 end fraction
  • fraction numerator open parentheses straight T subscript 1 plus straight T subscript 2 plus straight T subscript 3 close parentheses over denominator 3 end fraction

A.

fraction numerator straight n subscript 1 straight T subscript 1 plus straight n subscript 2 straight T subscript 2 space plus straight n subscript 3 straight T subscript 3 over denominator straight n subscript 1 plus space straight n subscript 2 plus space straight n subscript 3 end fraction

For adiabatic process i.e., no heat change

straight F over 2 straight n subscript 1 straight k subscript 1 straight T subscript 1 space plus fraction numerator begin display style straight F end style over denominator begin display style 2 end style end fraction straight n subscript 2 kT subscript 2 space plus fraction numerator begin display style straight F end style over denominator begin display style 2 end style end fraction straight n subscript 3 kT subscript 3 space
equals fraction numerator begin display style straight F end style over denominator begin display style 2 end style end fraction left parenthesis straight n subscript 1 plus straight n subscript 2 plus straight n subscript 3 right parenthesis kT
rightwards double arrow space straight T space equals space fraction numerator straight n subscript 1 straight T subscript 1 space plus straight n subscript 2 straight T subscript 2 plus straight n subscript 3 straight T subscript 3 over denominator space straight n subscript 1 plus straight n subscript 2 plus straight n subscript 3 end fraction

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7.

The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is

  • povo

  • open parentheses 13 over 2 close parentheses straight p subscript 0 straight v subscript 0
  • open parentheses 11 over 2 close parentheses straight p subscript straight o straight v subscript 0
  • 4povo


B.

open parentheses 13 over 2 close parentheses straight p subscript 0 straight v subscript 0

Heat is extracted from the source in path DA and AB is

space straight Q space equals space 3 over 2 straight R open parentheses fraction numerator straight P subscript 0 straight V subscript 0 over denominator straight R end fraction close parentheses space plus 5 over 2 straight R open parentheses fraction numerator begin display style 2 straight P subscript 0 straight V subscript 0 end style over denominator straight R end fraction close parentheses space equals space 13 over 2 straight P subscript 0 straight V subscript 0

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8.

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency

  • fraction numerator 1 over denominator 2 straight pi end fraction fraction numerator straight A subscript straight gamma straight p subscript 0 over denominator straight V subscript 0 straight M end fraction
  • fraction numerator 1 over denominator 2 straight pi end fraction fraction numerator straight V subscript 0 Mp subscript 0 over denominator straight A squared straight gamma end fraction
  • fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γp subscript 0 over denominator MV subscript 0 end fraction end root
  • fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator MV subscript 0 over denominator straight A subscript straight gamma straight p subscript 0 end fraction end root

C.

fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γp subscript 0 over denominator MV subscript 0 end fraction end root


FBD of piston at equilibrium
⇒ Patm A + mg = P0A

FBD of piston when piston is pushed down a distance x
straight P subscript atm space plus mg space minus space left parenthesis straight P subscript 0 plus dP right parenthesis space straight S space equals space straight m fraction numerator straight d squared straight x over denominator dt squared end fraction space left parenthesis ii right parenthesis
process space is space adiabatic space
rightwards double arrow space PV to the power of straight gamma space equals space straight C
rightwards double arrow straight C space equals dp equals space γPdV over straight V space space.. space left parenthesis iii right parenthesis
from space left parenthesis straight i right parenthesis space comma space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis
straight f space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of fraction numerator straight A squared γP subscript 0 over denominator MV subscript 0 end fraction end root

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9.

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion the average time of collision between molecules increases as Vq , where V is the volume of the gas. The value of q is: open parentheses straight gamma space equals space straight C subscript straight p over straight C subscript straight v close parentheses

  • fraction numerator 3 straight gamma plus 5 over denominator 6 end fraction
  • fraction numerator 3 straight gamma minus 5 over denominator 6 end fraction
  • fraction numerator straight gamma plus 1 over denominator 2 end fraction
  • fraction numerator straight gamma minus 1 over denominator 2 end fraction

C.

fraction numerator straight gamma plus 1 over denominator 2 end fraction

For an adiabatic process TVγ-1 = constant
We know that average time of collision between molecules

straight tau space equals space fraction numerator 1 over denominator nπ square root of 2 vrms end root straight d squared end fraction
Where n= number of molecules per unit volume

vrms = rms velocity of molecules

straight n space proportional to 1 over straight V space and space space straight v subscript rms space proportional to space square root of straight T
straight tau space proportional to fraction numerator straight V over denominator square root of straight T end fraction
Thus, we can write
n =K1V-1 and Vrms  = K2T1/2
Where K1 and K2 are constants.
For adiabatic process TVγ-1 = constant. Thus we can write
straight tau space proportional to space VT to the power of negative 1 divided by 2 end exponent space proportional to space straight V space left parenthesis straight V to the power of 1 minus straight gamma end exponent right parenthesis to the power of negative 1 divided by 2 end exponent
straight tau proportional to space straight V to the power of fraction numerator straight gamma plus 1 over denominator 2 end fraction end exponent

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10.

A Carnot engine operating between temperatures T1 and T2 has efficiency 1/6. When T2 is lowered by 62 K, its efficiency increases to 1/3. Then T1 and T2 are, respectively

  • 372 K and 330 K

  • 330 K and 268 K

  • 310 K and 248 K

  • 372 K and 310 K


D.

372 K and 310 K

The efficiency  is given by,

straight eta subscript 1 space equals space 1 space minus straight T subscript 2 over straight T subscript 1
rightwards double arrow space 1 over 6 space equals space 1 space minus space straight T subscript 2 over straight T subscript 1
rightwards double arrow space straight T subscript 2 over straight T subscript 1 space equals space 5 over 6
straight eta subscript 2 space equals space 1 minus fraction numerator straight T subscript 2 minus 62 over denominator straight T subscript 1 end fraction space space.... space left parenthesis straight i right parenthesis
1 third space equals space 1 minus fraction numerator straight T subscript 2 minus 62 over denominator straight T subscript 1 end fraction space... space left parenthesis ii right parenthesis
On space solving space Eqs space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
straight T subscript 1 space equals space 372 space straight K space and space straight T subscript 2 space equals space 310 space straight K

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