JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature $5 0^{\circ} C$ . The gas is now heated to double its temperature. The modified pressure is _____ Pa.

Your answer

Solution

NTA interprets “double its temperature” as doubling the Celsius value, i.e. from $5 0^{\circ}$ C to $10 0^{\circ}$ C (= 373 K).

For a fixed mass at constant volume, $P \propto T$ (in Kelvin):

$\frac{P _{f}}{P _{i}} = \frac{T _{f}}{T _{i}} \Rightarrow \frac{P _{f}}{3.23 kPa} = \frac{373}{323}$ .

$P_{f} = 3.23 \times \frac{373}{323} = 3.73$ kPa $= 3730$ Pa.

Note: A stricter reading would double the absolute temperature ( $323 \to 646$ K), giving $P_{f} = 6460$ Pa. The NTA-accepted answer follows the Celsius doubling.

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →