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 $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Heat & Thermodynamics

A container of fixed volume contains a gas at

2 7^{\circ} C

. To double the pressure of the gas, the temperature of gas should be raised to

____

^{\circ} C

Answer: 327

Solution

At constant volume, Gay-Lussac's law gives

\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}

With

T_{1} = 27° C = 300

K and

P_{2} = 2 P_{1}

\frac{P _{1}}{300} = \frac{2 P _{1}}{T _{2}} ⟹ T_{2} = 600 K

Converting back to Celsius,

T_{2} = 600 - 273 = 327° C

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Integer answer

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