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MCQEasyJEE 2025Equation of State of Ideal Gas

JEE Physics 2025 Question with Solution

A container of fixed volume contains a gas at 27∘C27^\circ \text{C}. To double the pressure of the gas, the temperature of the gas should be raised to _____°C.

  • A

    327∘C327^\circ \text{C}

  • B

    327 K327 \ \text{K}

  • C

    527∘C527^\circ \text{C}

  • D

    527 K527 \ \text{K}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A gas is kept in a container of fixed volume at 27∘C27^\circ \text{C}.

Find: The temperature required so that the pressure becomes double.

At constant volume, pressure is directly proportional to absolute temperature. So we use

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

Convert the initial temperature to Kelvin:

T1=27+273=300 KT_1 = 27 + 273 = 300 \, \text{K}

Since the final pressure is double,

P2=2P1P_2 = 2P_1

Substituting into the relation,

P1300=2P1T2\frac{P_1}{300} = \frac{2P_1}{T_2}

On simplifying,

T2=2×300=600 KT_2 = 2 \times 300 = 600 \, \text{K}

Convert back to Celsius:

T2=600−273=327∘CT_2 = 600 - 273 = 327^\circ \text{C}

Therefore, the temperature should be raised to 327∘C327^\circ \text{C}. The correct option is A.

Step-by-step Method

Given: Initial temperature is 27∘C27^\circ \text{C} and volume is fixed.

Find: Temperature at which pressure becomes twice its initial value.

For a fixed mass of gas at constant volume,

P∝TP \propto T

which gives

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

Because the pressure is doubled,

P2=2P1P_2 = 2P_1

Therefore,

P1T1=2P1T2\frac{P_1}{T_1} = \frac{2P_1}{T_2}

Cancelling P1P_1,

1T1=2T2\frac{1}{T_1} = \frac{2}{T_2}

So,

T2=2T1T_2 = 2T_1

Now convert 27∘C27^\circ \text{C} to Kelvin:

T1=27+273=300 KT_1 = 27 + 273 = 300 \, \text{K}

Hence,

T2=2×300=600 KT_2 = 2 \times 300 = 600 \, \text{K}

Finally, convert 600 K600 \, \text{K} to Celsius:

T2=600−273=327∘CT_2 = 600 - 273 = 327^\circ \text{C}

Thus, the required temperature is 327∘C327^\circ \text{C}.

Common mistakes

  • Using Celsius temperature directly in the gas law is incorrect because pressure is proportional to absolute temperature. Always convert ∘C^\circ \text{C} to K\text{K} first.

  • Assuming that doubling pressure means adding 27∘C27^\circ \text{C} again is wrong. The temperature must be doubled on the Kelvin scale, not on the Celsius scale.

  • Selecting 327 K327 \, \text{K} confuses the required unit. After solving in Kelvin, convert the result back to ∘C^\circ \text{C} because the blank asks for °C.

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