JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Heat & Thermodynamics

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One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is

100 K

and the universal gas constant

R = 8.0 J m o l^{- 1} K^{- 1}

, the decrease in its internal energy, in Joule, is ____.

Answer: 900

Solution

For an adiabatic process,

T V^{γ - 1} = constant

. With

γ = \frac{5}{3}

for a monatomic gas,

\frac{T _{f}}{T _{i}} = (\frac{V _{i}}{V _{f}})^{2/3} = (\frac{1}{8})^{2/3} = \frac{1}{4}

. So

T_{f} = \frac{100}{4} = 25

K. The change in internal energy is

Δ U = n C_{V} (T_{f} - T_{i}) = 1 \times \frac{3 R}{2} (25 - 100) = \frac{3 \times 8}{2} \times (- 75) = - 900

J. The decrease is 900 J.

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer

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