Pressure of an ideal gas, contained in a closed vessel, is increased by $0.4 \%$ when heated by $1^{\circ}…
- $25^{\circ} \mathrm{C}$
- $2500\mathrm{~K}$
- $250\mathrm{~K}$
- $250^{\circ} \mathrm{C}$
Solution
$\begin{aligned} & \mathrm{P} \propto \mathrm{T} \\ & \frac{\Delta \mathrm{P}}{\mathrm{P}}=\frac{\Delta \mathrm{T}}{\mathrm{T}}\end{aligned}$
$\begin{aligned} & \frac{0.4}{100}=\frac{1}{T} \\ & T=250 \mathrm{~K}\end{aligned}$
Asked in: JEE Main 2025 (03 Apr Shift 2)