Thermodynamics

JEE Main level, numerical answer, medium

80 of 190

Question 80 of 190

A container of fixed volume contains a gas at \(27^{\circ} \mathrm{C}\). To double the pressure of the gas, the temperature of gas should be raised to ______ \({ }^{\circ} \mathrm{C}\).

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Answer: 327

V= constant P1T1=P2T2P2=2P1T2=2T1=2×300=600 K∴T2=327∘C\begin{aligned} V & =\text { constant } \\ \frac{P_1}{T_1} & =\frac{P_2}{T_2} \\ P_2 & =2 P_1 \\ T_2 & =2 T_1 \\ & =2 \times 300=600 \mathrm{~K} \\ \therefore T_2 & =327^{\circ} \mathrm{C}\end{aligned}

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