Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, the net work done by the gas is

Question

Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, the net work done by the gas is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ade16e76627dc3d0f2138d31cfc48837-.png" /> (A)  200 R In 2  (B)  100 R In 2  (C)  300 R In 2  (D)  400 R In 2

Tardigrade Admin · Accepted Answer

Process AB and CD - isobaric Workdone = P Δ V = nR Δ T WAB = nR (100) WCD =- nR (100) Process BC and DA ⇒ isothermal Work done = nRT ln ( P 1/ P 2) WBC = nRT ln (1/2)=-400 nR ln (1/2) W Δ A = nRT ln 2=300 nR ln 2 Total work done =100 nR -100 nR +400 R ln 2-300 nR ln 2 =100 nR ln 2 =200 R ln 2 ( n =2)

Q. Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, the net work done by the gas is

$200 R I n 2$

$100 R I n 2$

$300 R I n 2$

$400 R I n 2$

Q. Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, the net work done by the gas is

200RIn2

100RIn2

300RIn2

400RIn2

Process AB and CD− isobaric Workdone =PΔV=nRΔT WAB=nR(100) WCD=−nR(100) Process BC and DA⇒ isothermal Work done =nRTlnP2​P1​​ WBC =nRTln21​=−400nRln21​ W ΔA=nRTln2=300nRln2 Total work done =100nR−100nR+400Rln2−300nRln2 =100nRln2 =200Rln2(n=2)

$200 R I n 2$

$100 R I n 2$

$300 R I n 2$

$400 R I n 2$