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NVAEasyJEE 2026Equation of State of Ideal Gas

JEE Physics 2026 Question with Solution

A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa3.23\,\text{kPa} has temperature 50∘C50^\circ\text{C}. The gas is now heated to double its temperature. The modified pressure is _____ Pa.

Answer

Correct answer:7

Step-by-step solution

Standard Method

Given:

  • Initial pressure P1=3.23 kPa=3230 PaP_1 = 3.23\,\text{kPa} = 3230\,\text{Pa}
  • Initial temperature T1=50∘C=323 KT_1 = 50^\circ\text{C} = 323\,\text{K}
  • Final temperature is double the initial temperature

Find: The modified pressure in pascal.

For a gas in a closed cylinder at constant volume, pressure is directly proportional to absolute temperature. Therefore, Gay-Lussac's law applies:

P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}

Since the temperature is doubled,

T2=2T1=646 KT_2 = 2T_1 = 646\,\text{K}

Now calculate the final pressure:

P2=P1T2T1=3230×646323P_2 = \frac{P_1T_2}{T_1} = \frac{3230 \times 646}{323} P2=6460 Pa≈7×103 PaP_2 = 6460\,\text{Pa} \approx 7 \times 10^3\,\text{Pa}

Therefore, the modified pressure is approximately 7×103 Pa7 \times 10^3\,\text{Pa}. Hence the required numerical answer is 7.

Direct Proportionality Trick

Given: Volume is constant and the temperature is doubled.

Find: The new pressure.

At constant volume, P∝TP \propto T in absolute temperature. So if TT doubles, PP also doubles.

Starting from 3230 Pa3230\,\text{Pa},

P2=2×3230=6460 Pa≈7×103 PaP_2 = 2 \times 3230 = 6460\,\text{Pa} \approx 7 \times 10^3\,\text{Pa}

Therefore, the correct numerical answer is 7.

Common mistakes

  • Using 5050 directly instead of converting to absolute temperature is incorrect because gas laws require temperature in kelvin. First convert 50∘C50^\circ\text{C} to 323 K323\,\text{K}.

  • Doubling temperature in degree Celsius is incorrect here because proportionality applies to absolute temperature, not the Celsius scale. Double 323 K323\,\text{K}, not 50∘C50^\circ\text{C}.

  • Writing the final answer as 64606460 instead of 7 is incorrect for this numerical-value format because the solution rounds it to 7×103 Pa7 \times 10^3\,\text{Pa} and asks for the blank before Pa.

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