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Question

An ideal gas whose adiabatic exponent equals to γ=75 is expanded according to the law P=2V. The initial volume of the gas is equal to V0=1 unit. As a result of expansion volume increases 4 times. (Take R=253 units )

Column I Column II
(A) Work done by the gas (P)25 units
(B) Increment in internal energy of the gas (Q) 90 units
(C) Heat supplied to the gas (R) 75 units
(D) Molar heat capacity of the gas in the process (S) 15 units
(D) 55 units


A
(A)(S);(B)(R);(C)(Q);(D)(P)
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B
(A)(P);(B)(Q);(C)(R);(D)(T)
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C
(A)(T);(B)(S);(C)(P);(D)(S)
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D
(A)(S);(B)(R);(C)(P);(D)(Q)
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Solution

The correct option is A (A)(S);(B)(R);(C)(Q);(D)(P)
Given,
γ=75,P=2V,Vo=1 unit
and Vf=4V0

Work done in an adiabatic process W=VfViPdV
W=4V0V02VdV=[V2]4V0V0=15V20=15 units
Hence, (A)(S).

Increment in internal energy
Ideal gas equation : PV=nRT,
where, P=2V
so, 2V2=nRT2(V2fV20)=nR(ΔT)2(4212)V20=nR(ΔT)nRΔT=30V20 ....(1)

Internal energy in an adiabatic process is given by
ΔU=nCvΔT=nRγ1ΔT
using eq. (1),
30V20γ1=30(1)2751=302(5)=75 units
Hence, (B)(R).

Calculation of heat supply,
Q=W+ΔU
As we have calculated early,
W=15 unit
and ΔU=75 unit
Q=15+75=90 units
Hence, Hence, (C)(Q).

Molar heat capacity for PVx process is given as:
C=Cv+R1x=52R+R1(1)=52R+R2=3R=3×253=25 units
Hence, Hence, (D)(P).

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