Q 1 :    

A sample of gas at temperature T is adiabatically expanded to double its volume. Adiabatic constant for the gas is γ=3/2. The work done by the gas in the process is (n = 1 mole)               [2024]

  • RT[1-22]

     

  • RT[22-1]

     

  • RT[2-2]

     

  • RT[2-2]

     

(D)    PV=nRTP=nRTV

          Also, PVγ= constant

         nRTVVγ= constant TVγ-1= constant

          TVγ-1=T2·(2V)γ-1

           T2=T·(12)1/2=T2

            W=nR(T1-T2)γ-1=2nR(T-T2)=RT(2-2)

          

 



Q 2 :    

During an adiabatic process, if the pressure of a gas is found to be proportional to the cube of its absolute temperature, then the ratio of CPCV for the gas is                          [2024]

  • 32

     

  • 75

     

  • 53

     

  • 97

     

(A)   P=T3PT-3= constant

         PVT=nR= constant from ideal gas equation T=PVnR

         P(PVnR)-3= constant P-2V-3=constant

        PV3/2= constant                                   ... (1)

          Process equation for adiabatic process is 

          PVγ= constant                                           ... (2)

         Comparing equation (1) and (2)

         CPCV=γ=32

 



Q 3 :    

A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2°C. The work done by the gas is (Given: R=8.31 J k-1 mol-1)                 [2024]

  • 48 J

     

  • 23.1 J

     

  • 24.9 J

     

  • 72.9 J

     

(B)   Given : Q=48 J

         1st law of thermodynamics, Q=U+W

          48=nCvT+W

           48=(1)(3R2)(2)+W

           W=48-3×R

           W=48-3×(8.3)       

             W=23.1 J

 



Q 4 :    

A diatomic gas (γ=1.4) does 100 J of work in an isobaric expansion. The heat given to the gas is                     [2024]

  • 150 J

     

  • 250 J

     

  • 490 J

     

  • 350 J

     

(D)   W=100 J

        Q=nCpT=n(7R2)T=72nRT

        PV=nRTPV=nRT= work done (W)

        Q=72nRT=72W

         Q=72×100=350 J

 



Q 5 :    

The volume of an ideal gas (γ=1.5) is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is:           [2024]

  • 45

     

  • 1625

     

  • 855

     

  • 25

     

(C)   For adiabatic process, PVγ=k

         P1V13/2=P2V23/2

          P1×(5)3/2=P2×(4)3/2

          P1P2=(45)3/2=(2)3(125)1/2=855

 



Q 6 :    

A sample of 1 mole gas at temperature T is adiabatically expanded to double its volume. If adiabatic constant for the gas is γ=32, then the work done by the gas in the process is:                 [2024]

  • RT[2-2]

     

  • RT[2-2]

     

  • RT[2+2]

     

  • TR[2+2]

     

(A)     WD by gas in adiabatic process

           W=P2V2-P1V11-γ=nR[T2-T1]1-γ

           T1V1γ-1=T2V2γ-1

            T×(V)1/2=T2×(2V)1/2T2=T2

             W=R[T2-T]1-32=2R2[2T-T]

              W=RT[2-2]

 



Q 7 :    

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of Cp/Cv for the gas is                         [2024]

  • 53

     

  • 32

     

  • 75

     

  • 97

     

(B)     PT3PT-3= constant

            PVγ= const

           P(nRTP)γ= const

           P1-γTγ= const PTγ1-γ= const

            γ1-γ=-3

             γ=-3+3γγ=32

 



Q 8 :    

The pressure and volume of an ideal gas are related as PV3/2=K (Constant). The work done when the gas is taken from state A(P1,V1,T1) to state B(P2,V2,T2) is                 [2024]

  • 2(P1V1-P2V2)

     

  • 2(P2V2-P1V1)

     

  • 2(P1V1-P2V2)

     

  • 2(P2V2-P1V1)

     

(A)    For PVx= constant

         W=nRT1-x

          Here x=32

                W=P2V2-P1V1-12

          =2(P1V1-P2V2)



Q 9 :    

A diatomic gas (γ=1.4) does 200 J of work when it is expanded isobarically. The heat given to the gas in the process is               [2024]

  • 850 J

     

  • 800 J

     

  • 600 J

     

  • 700 J

     

(D)    Q=PV+nCVT

               =PV+nR(γ-1)T=PV+PVγ-1

               =PV(1+1γ-1)

                =200(1+10.4)

                200×3.5=700 J