QJEEAdv18Ph01d2: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph01d2 Solution: One mole of a monatomic ideal gas undergoes an adiabatic…

The degrees of freedom of a monatomic gas is f = 3. Its molar specific heats are ${C_v} = {f \over 2}RT$ and ${C_p} = {C_v} + R = {{(f + 2)} \over 2}RT$ . The ratio of specific heats is given by

$\gamma = {C_p}/{C_v} = (f + 2)/f = 5/3$ .

In an adiabatic process, $TV_{}^{\gamma - 1}$ = constant. Hence, ${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$ , which gives

${T_2} = {T_1}{({V_1}/{V_2})^{\gamma - 1}} = 100{(1/8)^{5/3 - 1}} = 25\,K$ ,

where we used T₁ = 100 K and V₂ = 8V₁. The change in internal energy of one mole of the ideal gas is

$\Delta U = {U_2} - {U_1} = {f \over 2}R{T_2} - {f \over 2}R{T_1} = {f \over 2}R({T_2} - {T_1})$

$= (3/2)(8)(25 - 100) = - 900\,J$ .

Thus, the decrease in internal energy of the gas is 900 J.