The temperature ofof$\mathbf{3}\mathbf{.}\mathbf{00}\mathit{m}\mathit{o}\mathbf{\text{l}}$an ideal diatomic gas is increased by $\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$without the pressure of the gas changing. The molecules in the gas rotate but do not oscillate.

a) How much energy is transferred to the gas as heat?

b) What is the change in the internal energy of the gas?

c) How much work is done by the gas?

d) By how much does the rotational kinetic energy of the gas increase?

• Number of moles$n=3\text{\hspace{0.17em}mol}$
• Temperature is$T=40°\text{C}$

The expression for the amount of energy transferred to the gas is given by,

$\mathit{Q}\mathbf{=}\mathit{n}{\mathit{C}}_{\mathbf{P}}\mathit{\Delta }\mathit{T}$

Here$Q$ is the amount of energy transferred to gas, $n$is the number of moles, $C_p$is the specific heat capacity at constant pressure and $\Delta T$is the temperature difference.

The expression for the specific heat capacity is given by,

$C_p=\frac{7}{2}R$

Here$R$ is the gas constant.

Energy transferred to the gas as heat:

For a diatomic gas,
$C_V=\frac{5}{2}R$and as$C_P=C_V+R$,

$\begin{array}{c}{C}_P=\frac{5}{2}R+R\\ C_P=\frac{7}{2}R\end{array}$

The molar specific heat$C_P$is

$\begin{array}{c}{C}_P=\frac{Q}{n\Delta T}\\ Q=n{C}_P\Delta T\end{array}$

Where,$Q$is the energy transferred as heat to or from a sample of$n$mole of the gas,$\Delta T$is the resulting change in temperature of the gas, and$\Delta E_{int}$is the change intheinternal energy of the gas.

Substituting$\frac{7}{2}R$ for$8.314J/K\cdot mol$for$R$ , for $40°C$, for $\Delta T$and $3$for$n$ into the above equation.

$\begin{array}{c}Q=3\times \frac{7}{2}R\times \Delta T\\ =3\times \frac{7}{2}\times 8.31\times 40\\ =3.493\times {10}^3\text{J}\end{array}$

Therefore the energy transferred to the gas as heat is$3.49\times {10}^3\text{J}$ .

The molar specific heat ${\text{C}}_V$of a gas at constant volume is given as

$\begin{array}{c}{C}_V=\frac{Q}{n\Delta T}\\ =\frac{\Delta E_{int}}{n\Delta T}\end{array}$

Thus,$\Delta E_{int}=n{C}_V\Delta T$

Also,

$C_V=\frac{5}{2}R$

Substituting$\frac{5}{2}R$ for $C_v$,$8.314J/K\cdot mol$ for $R$, $40°C$for$\Delta T$ and $3$fordata-custom-editor="chemistry" $\mathrm{n}$ into the equation of change in internal energy.

$\begin{array}{c}\Delta E_{int}=n\times \frac{5}{2}R\times 40\\ =3\times \frac{5}{2}\times 8.31\times 40\\ =2.493\times {10}^3\text{J}\end{array}$

Therefore the change in internal energy of the gas is $2.49\times {10}^3\text{J}$.

Using first law of thermodynamics,

$\begin{array}{c}Q=\Delta E_{int}+W\\ W=Q-\Delta E_{int}\end{array}$

Substitute $3.49\times {10}^3\text{J}$for$Q$and $2.49\times {10}^3\text{J}$ for$\Delta E_{int}$ into the above equation.

$\begin{array}{c}W=3.49\times {10}^3\text{J}-2.49\times {10}^3\text{J}\\ =(3.49-2.493)\times {10}^3\text{J}\\ =9.97\times {10}^2\text{J}\end{array}$

Therefore the work done by the gas is$9.97\times {10}^2\text{J}$ .

The average translational kinetic energy ofmolecules of gas is given as

$\begin{array}{c}{K}_{avg}=\frac{3}{2}nKT\\ K_{avg}=\frac{3}{2}RT\end{array}$

Therefore, the increase inthetranslational kinetic energy is given as

$\begin{array}{c}\Delta K_{trans}=\text{Δ}\left(n{K}_{avg}\right)\\ =n\frac{3}{2}R\Delta T\end{array}$

Substituting the values for gas constant, change in temperature, and number of moles, we get

$\begin{array}{c}\Delta K_{trans}=3\times \frac{3}{2}\times 8.31\times 40\\ =1.49\times {10}^3\text{J}\end{array}$

Since,${\text{ΔE}}_{int}={\text{ΔK}}_{trans}+{\text{ΔK}}_{rot}$

$\begin{array}{c}\Delta K_{rot}=\Delta E_{int}-\Delta K_{trans}\\ =2.493\times {10}^3\text{J}-1.49\times {10}^3\text{J}\\ =1.003\times {10}^3\text{J}\\ \approx 1\times {10}^3\text{J}\end{array}$

Therefore Increase in the rotational kinetic energy of the gas is$1\times {10}^3\text{J}$