Thermodynamics

JEE Main level, single correct, medium

164 of 190

Question 164 of 190

During an adiabatic compression, 830 J830 \mathrm{~J} of work is done on 2 moles of a diatomic ideal gas to reduce its volume by 50%50 \%. The change in its temperature is nearly: (R=8.3JK−1 mol−1)\left(\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)

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Answer: (C) 20 K20 \mathrm{~K}

Given : work done, W=830 JW=830 \mathrm{~J}
No. of moles of gas, μ=2\mu=2
For diatomic gas γ=1.4\gamma=1.4
Work done during an adiabatic change
W=μR(T1−T2)γ−1⇒830=2×8.3(ΔT)1.4−1=2×8.3(ΔT)0.4⇒ΔT=830×0.42×8.3=20 K\begin{aligned} &W=\frac{\mu R\left(T_1-T_2\right)}{\gamma-1} \\ &\Rightarrow 830=\frac{2 \times 8.3(\Delta T)}{1.4-1}=\frac{2 \times 8.3(\Delta T)}{0.4} \\ &\Rightarrow \Delta T=\frac{830 \times 0.4}{2 \times 8.3}=20 \mathrm{~K} \end{aligned}

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