Thermodynamics

JEE Main level, single correct, medium

176 of 190

Question 176 of 190

A perfect gas at 27∘C27^{\circ} \mathrm{C} is heated at constant pressure so as to double its volume. The final temperature of the gas will be, close to

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Answer: (A) 327∘C327^{\circ} \mathrm{C}

Given, V1=VV_1=V
V2=2VT1=27∘+273=300 KT2=?\begin{aligned} & V_2=2 V \\ & T_1=27^{\circ}+273=300 \mathrm{~K} \\ & T_2=? \end{aligned}
From charle's law

V1T1=V2T2(∵ Pressure is constant )\frac{V_1}{T_1}=\frac{V_2}{T_2}(\because \text { Pressure is constant })
or,
V300=2VT2\frac{V}{300}=\frac{2 V}{T_2}

∴T2=600 K=600−273=327∘C\therefore \quad T_2=600 \mathrm{~K}=600-273=327^{\circ} \mathrm{C}

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