Physics Q34 – JEE-MAIN 2025

Pressure of an ideal gas, contained in a closed vessel, is increased by

0.4 %

when heated by

1^{\circ} C

. Its initial temperature must be:

Clue 1 of 3: Ideal Gas Behavior

For an ideal gas in a closed vessel, the pressure is directly proportional to its absolute temperature.

🥷

Ninja Strategy•Approximate Proportionality Check

Recognize that for small changes, the fractional change in pressure is approximately equal to the fractional change in absolute temperature ( $\frac{Δ P}{P} \approx \frac{Δ T}{T}$ ), allowing for quick estimation and elimination of options with significantly different magnitudes.

🧠 Full Solution Path

Step 1: Identify the Governing Principle✦ Active

For an ideal gas in a closed vessel, the volume ( $V$ ) and the number of moles ( $n$ ) are constant. According to the Ideal Gas Law ( $P V = n R T$ ), this implies that pressure ( $P$ ) is directly proportional to the absolute temperature ( $T$ ). Thus, we can write $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ .

Step 2: Express Given Changes in Terms of Initial Conditions○ Expand

Let the initial pressure be $P_{1}$ and the initial absolute temperature be $T_{1}$ . Given that the pressure increases by $0.4 %$ when heated by $1^{\circ} C$ (which is $1 K$ in absolute temperature change):

P_{2} = P_{1} + 0.004 P_{1} = 1.004 P_{1}

T_{2} = T_{1} + 1 K

Step 3: Solve for the Initial Temperature○ Expand

Substitute the expressions for $P_{2}$ and $T_{2}$ into the proportionality equation and solve for $T_{1}$ :

\frac{P_{1}}{T_{1}} = \frac{1.004 P_{1}}{T_{1} + 1}

T_{1} + 1 = 1.004 T_{1}

1 = 1.004 T_{1} - T_{1}

1 = 0.004 T_{1}

T_{1} = \frac{1}{0.004} = 250 K

The initial temperature is $250 K$ .

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 34 – JEE-MAIN 2025

Clue 1 of 3: Ideal Gas Behavior

🧠 Full Solution Path