Question 116: Thermodynamics & KTG

Question

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 $$R=8.0$$ $$J\,mo{l^{ - 1}}{K^{ - 1}},$$ the decrease in its internal energy, in Joule, is ____________.

Accepted Answer

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 T
1
= 100 K and V
2
= 8V
1
. 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.

Thermodynamics & KTG

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