Question 1: 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

Integer / Numeric Answer Type