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Q.

Two moles of ideal gas Cv,m=52Rwas compressed adiabatically against constant pressure of 2atm which was initially at 350K and 1atm the work done on the gas in this process is

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a

250R

b

500R

c

125R

d

300R

answer is B.

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Detailed Solution

ΔU=Q+WΔU=W(∵q=0 in adiabatic process )W=nCvΔT=nCvTf-TiW=-PextVf-Vi-PfnRTfPf-nRTipi=n×52RTf-Ti-Tf-TiPfPi=52Tf-Ti-Tf-350×21=52Tf-350⇒Tf=450K Now W=nCvmTf-Ti  =2×52R(450-350)=500R
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Two moles of ideal gas Cv,m=52Rwas compressed adiabatically against constant pressure of 2atm which was initially at 350K and 1atm the work done on the gas in this process is