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MCQEasyJEE 2026Isothermal & Adiabatic Processes

JEE Physics 2026 Question with Solution

The volume of an ideal gas increases 88 times and temperature becomes (14)th\left(\frac{1}{4}\right)^{\text{th}} of initial temperature during a reversible change. If there is no exchange of heat in this process (ΔQ=0)(\Delta Q=0), then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

  • A

    He

  • B

    O2\mathrm{O_2}

  • C

    CO2\mathrm{CO_2}

  • D

    NH3\mathrm{NH_3}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The volume becomes V2=8V1V_2 = 8V_1, the temperature becomes T2=14T1T_2 = \frac{1}{4}T_1, and there is no exchange of heat, so (ΔQ=0)(\Delta Q = 0) for a reversible process.

Find: Identify the gas.

Since there is no exchange of heat and the process is reversible, the process is adiabatic.

For a reversible adiabatic process of an ideal gas:

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

Step-by-step Evaluation

Apply the given data:

T1V1γ−1=T2V2γ−1T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1} T1V1γ−1=14T1(8V1)γ−1T_1V_1^{\gamma-1} = \frac{1}{4}T_1(8V_1)^{\gamma-1}

Canceling T1V1γ−1T_1V_1^{\gamma-1} gives:

1=14 8γ−11 = \frac{1}{4} \, 8^{\gamma-1}

So,

8γ−1=48^{\gamma-1} = 4

Write both sides as powers of 22:

(23)γ−1=22(2^3)^{\gamma-1} = 2^2 3(γ−1)=23(\gamma-1)=2 γ=53\gamma = \frac{5}{3}

The ratio of specific heats γ=53\gamma = \frac{5}{3} corresponds to a monoatomic gas.

Among the given options, only Helium (He) is monoatomic.

Therefore, the correct option is A.

Common mistakes

  • Mistake: Using the isothermal relation instead of the adiabatic relation. This is wrong because the question states (ΔQ=0)(\Delta Q=0) and the process is reversible, which means it is adiabatic. Use TVγ−1=constantTV^{\gamma-1}=\text{constant} instead.

  • Mistake: Identifying the gas only from molecular formula familiarity without calculating γ\gamma. This is wrong because the gas must be inferred from the thermodynamic relation first. Find γ\gamma and then match it to the nature of the gas.

  • Mistake: Solving 8γ−1=48^{\gamma-1}=4 incorrectly by not expressing both numbers in the same base. This leads to a wrong value of γ\gamma. Write 8=238=2^3 and 4=224=2^2 before comparing powers.

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