Kinetic Theory

JEE Main level, single correct, medium

68 of 168

Question 68 of 168

The volume of an ideal gas increases 8 times and temperature becomes (1/4)th (1 / 4)^{\text {th }} of initial temperature during a reversible change. If there is no exchange of heat in this process (ΔQ=0)(\Delta \mathrm{Q}=0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):

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Answer: (D) He

For an adiabatic process (ΔQ=0\Delta Q = 0) with an ideal gas: TVγ−1=constantTV^{\gamma-1} = \text{constant}
Given: Vf=8ViV_f = 8V_i and Tf=14TiT_f = \frac{1}{4}T_i
Applying the adiabatic relation:
TiViγ−1=TfVfγ−1T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}
TiViγ−1=Ti4(8Vi)γ−1T_i V_i^{\gamma-1} = \frac{T_i}{4}(8V_i)^{\gamma-1}
1=14⋅8γ−11 = \frac{1}{4} \cdot 8^{\gamma-1}
4=8γ−14 = 8^{\gamma-1}
22=23(γ−1)2^2 = 2^{3(\gamma-1)}
2=3(γ−1)2 = 3(\gamma-1)
γ=53=1.67\gamma = \frac{5}{3} = 1.67
This value corresponds to a monatomic gas. Among the options, only He is monatomic (γ=5/3\gamma = 5/3 for monatomic ideal gases). O₂ and N₂ are diatomic (γ=7/5\gamma = 7/5), CO₂ is polyatomic (γ≈1.30\gamma \approx 1.30), and NH₃ is polyatomic.

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