Thermodynamics

JEE Main level, single correct, medium

153 of 190

Question 153 of 190

Two moles of an ideal monoatomic gas occupies a volume VV at 27oC{27}^{o}C . The gas expands adiabatically to a volume 2V2V. Calculate (a)(a) the final temperature of the gas and (b)(b) change in its internal energy.

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Answer: (D) (a)189K(b)−2.7kJ(a)189K(b)-2.7kJ

TVγ−1=ConstantT{V}^{\gamma -1}=Constant
γ=53\gamma =\dfrac{5}{3}⇒300(V)23=T(2V)23\Rightarrow 300{\left(V\right)}^{\dfrac{2}{3}}=T{\left(2V\right)}^{\dfrac{2}{3}}
⇒T=189K\Rightarrow T=189K⇒ΔU=nCvΔT\Rightarrow \Delta U=n{C}_{v}\Delta T=2×32×8.314×(−11)=2\times \dfrac{3}{2}\times 8.314\times \left(-11\right)=−2.7kJ=-2.7kJ

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