JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Heat & Thermodynamics

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A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The rational between temperature and volume for the process is TV

^{x} =

constant, then x is:

(A) $3/5$
(B) $2/5$ ✓ Correct
(C) $2/3$
(D) $5/3$

Answer: (B)

Solution

For an adiabatic change

T V^{γ - 1} = constant

, so comparing with

T V^{x} = constant

gives

x = γ - 1

. A rigid diatomic molecule at room temperature has three translational and two rotational degrees of freedom,

f = 5

, with vibration frozen out, so

γ = \frac{f + 2}{f} = \frac{7}{5}

. Hence

x = \frac{7}{5} - 1 = \frac{2}{5}

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option

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