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JEE Physics 2026 Question with Solution

One mole of ideal diatomic gas expands... final temperature will be (close to) _____ ∘C^\circ\text{C}.

  • A

    −56-56

  • B

    −30-30

  • C

    −189-189

  • D

    −117-117

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: One mole of an ideal diatomic gas undergoes adiabatic expansion. Find: the final temperature in ∘C^\circ\text{C}.

For an adiabatic process,

PVγ=constantPV^\gamma = \text{constant}

For a diatomic gas,

γ=75\gamma = \frac{7}{5}

Also, the temperature-volume relation is

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

From the solution text, the volume doubles, so

V2=2V1V_2 = 2V_1

Therefore,

T2T1=(12)0.4≈0.742\frac{T_2}{T_1} = \left(\frac{1}{2}\right)^{0.4} \approx 0.742

Hence,

T2=0.742 T1T_2 = 0.742 \, T_1

Using the value taken in the solution,

T1≈300 KT_1 \approx 300 \, \text{K}

So,

T2=300×0.742=222.6 KT_2 = 300 \times 0.742 = 222.6 \, \text{K}

Converting to Celsius,

T2≈222.6−273=−50.4 ∘CT_2 \approx 222.6 - 273 = -50.4 \, ^\circ\text{C}

The closest option is −56-56. Therefore, the correct option is A.

Common mistakes

  • Using the isothermal relation instead of the adiabatic relation is incorrect because in an adiabatic process temperature changes. Use PVγ=constantPV^\gamma = \text{constant} or T2T1=(V1V2)γ−1\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} instead.

  • Taking the wrong value of γ\gamma for a diatomic gas gives an incorrect temperature drop. For an ideal diatomic gas, use γ=75\gamma = \frac{7}{5}, not 53\frac{5}{3}.

  • Forgetting to convert Kelvin to Celsius at the end leads to a wrong option choice. First compute the final temperature in Kelvin, then subtract 273273 to express it in ∘C^\circ\text{C}.

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