Thermodynamics

JEE Main level, single correct, medium

157 of 190

Question 157 of 190

The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is

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Answer: (A) 25\dfrac{2}{5}

Work done by an ideal daitomic gas is given by, W=∆Q−∆U∆Q=nCp∆TwhereCpfordiatomicgasis52R∆U=nCv∆TwhereCvfordiatomicgasis32RhenceW=n52R∆T−n32R∆T=nR∆T(52−32)=nR∆TW=∆Q-∆U ∆Q=n{C}_{p}∆Twhere{C}_{p}fordiatomicgasis\dfrac{5}{2}R ∆U=n{C}_{v}∆Twhere{C}_{v}fordiatomicgasis\dfrac{3}{2}R henceW=n\dfrac{5}{2}R∆T-n\dfrac{3}{2}R∆T=nR∆T\left(\dfrac{5}{2}-\dfrac{3}{2}\right)=nR∆T
And heat supplies is given by, (∆Q)p=nCp∆T=n52R∆T(∆Q{)}_{p}=n{C}_{p}∆T=n\dfrac{5}{2}R∆T
Hence the ratio is given by∆W(∆Q)P=nR∆Tn52R∆T\dfrac{∆W}{{\left(∆Q\right)}_{P}}=\dfrac{nR∆T}{n\dfrac{5}{2}R∆T}

∆W(∆Q)P=25\dfrac{∆W}{{\left(∆Q\right)}_{P}}=\dfrac{2}{5}

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