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MCQEasyJEE 2025Isothermal & Adiabatic Processes

JEE Physics 2025 Question with Solution

A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800 cm3800 \, \text{cm}^3 and temperature 27∘C27^\circ \text{C}. The change in temperature when the gas is adiabatically compressed to 200 cm3200 \, \text{cm}^3 is: (Take γ\gamma = 1.51.5 : γ\gamma is the ratio of specific heats at constant pressure and at constant volume)

  • A

    327 K327 \, \text{K}

  • B

    600 K600 \, \text{K}

  • C

    522 K522 \, \text{K}

  • D

    300 K300 \, \text{K}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Initial volume V1=800 cm3V_1 = 800 \, \text{cm}^3, final volume V2=200 cm3V_2 = 200 \, \text{cm}^3, initial temperature T1=27∘CT_1 = 27^\circ \text{C}, and γ=1.5\gamma = 1.5.

Find: The change in temperature during adiabatic compression.

Since the walls are thermally non-conducting, the process is adiabatic. For a reversible adiabatic process,

TVγ−1=constantT V^{\gamma-1} = \text{constant}

So, for the two states,

T1V1γ−1=T2V2γ−1T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}

Convert the initial temperature to Kelvin:

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

Now solve for the final temperature:

T2=T1(V1V2)γ−1T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1}

Substitute the given values:

T2=300(800200)1.5−1T_2 = 300 \left(\frac{800}{200}\right)^{1.5-1} T2=300(4)0.5=300×2=600 KT_2 = 300 (4)^{0.5} = 300 \times 2 = 600 \, \text{K}

The question asks for the change in temperature:

ΔT=T2−T1=600−300=300 K\Delta T = T_2 - T_1 = 600 - 300 = 300 \, \text{K}

Therefore, the change in temperature is 300 K300 \, \text{K} and the correct option is D.

Direct Ratio Method

Given: V1=800 cm3V_1 = 800 \, \text{cm}^3, V2=200 cm3V_2 = 200 \, \text{cm}^3, T1=300 KT_1 = 300 \, \text{K}, γ=1.5\gamma = 1.5.

Find: Temperature change ΔT\Delta T.

Use the adiabatic relation directly:

T2T1=(V1V2)γ−1\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}

Here,

V1V2=4,γ−1=0.5\frac{V_1}{V_2} = 4, \qquad \gamma - 1 = 0.5

So,

T2300=41/2=2\frac{T_2}{300} = 4^{1/2} = 2

Hence,

T2=600 KT_2 = 600 \, \text{K}

Therefore,

ΔT=600−300=300 K\Delta T = 600 - 300 = 300 \, \text{K}

The correct option is D.

Common mistakes

  • Using the final temperature T2=600 KT_2 = 600 \, \text{K} as the answer. This is wrong because the question asks for the change in temperature, not the final temperature. Always compute ΔT=T2−T1\Delta T = T_2 - T_1.

  • Not converting 27∘C27^\circ \text{C} to Kelvin before applying the adiabatic relation. This is wrong because thermodynamic temperature relations require the absolute scale. First take T1=300 KT_1 = 300 \, \text{K}.

  • Using the wrong exponent in the adiabatic formula, such as γ\gamma instead of γ−1\gamma - 1. This gives an incorrect temperature ratio. For temperature-volume relation, use TVγ−1=constantT V^{\gamma-1} = \text{constant}.

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