Under constant pressure, the temperature of$\mathbf{2}\mathbf{.}\mathbf{00}\mathit{m}\mathit{o}\mathbf{\text{l}}$of an ideal monoatomic gas is raised$\mathbf{15}\mathbf{.}\mathbf{0}\mathit{K}$. What are

1. The work$\mathit{W}$ done by the gas
2. The energy transferred as heat$\mathit{Q}$
3. The change$\mathit{\Delta }{\mathit{E}}_{\mathbf{i}\mathbf{n}\mathbf{t}\mathbf{}}$in the internal energy of the gas
4. The change$\mathit{\Delta }\mathit{K}$in the average kinetic energy per atom?

• Change in temperature is$\Delta T=15.0\text{\hspace{0.17em}K}$
• The number of moles of gas$n=2.00\text{\hspace{0.17em}mol}$

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

$Q=n{C}_P\Delta T$

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 work done by the gas is given by,

$W=p\Delta V$

Here $W$is the work done by the gas, $p$is the pressure and$\Delta V$ is the change in the volume.

Work done by the gas$W$;

Since the process is a constant-pressure

$W=p\Delta V$

Using ideal gas equation,

$\begin{array}{c}W=nR\Delta T\\ =\left(2\text{mol}\right)\left(8.31\frac{\text{J}}{\text{mol}.\text{K}}\right)\left(15\text{\hspace{0.17em}K}\right)\\ =249\text{\hspace{0.17em}J}\end{array}$

Therefore the work done by the gas is$249\text{\hspace{0.17em}J}$ .

The energy transferred as heat$Q$;

For mono-atomic gas$f=3$ ,

$C_V=\frac{3}{2}R$

${\text{C}}_P={\text{C}}_V+\text{R}$

Substitute$\frac{3}{2}R$ for$C_V$ into the above equation,

$\Rightarrow C_P=\frac{5}{2}R$

The molar specific heat$C_P$of a gas is defined as

$C_P=\frac{Q}{n\Delta T}$

Solving for$\text{Q,}$we get

$Q=n{C}_P\Delta T$

$\begin{array}{c}Q=\left(2\text{mol}\right)\left(\frac{5}{2}\right)\left(8.31\frac{\text{J}}{\text{mol.K}}\right)\left(15\text{\hspace{0.17em}K}\right)\\ =623\text{\hspace{0.17em}J}\end{array}$

Therefore the energy transferred as heat$Q$ is $623\text{\hspace{0.17em}J}$.

The change intheinternal energy of the gas$\Delta E_{int}$;

Fromthefirst law of thermodynamics,

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

Substitute$623\text{\hspace{0.17em}J}$ for$Q$ and$249\text{\hspace{0.17em}J}$ for$W$ into the above equation.

$\begin{array}{c}\Delta E_{int}=623\text{\hspace{0.17em}J}-249\text{\hspace{0.17em}J}\\ =374\text{\hspace{0.17em}J}\end{array}$

Therefore the change in the internal energy of the gas is $374\text{\hspace{0.17em}J}$.

The change intheaverage kinetic energy per atom $\Delta K_{avg}$;

The change in average kinetic energy per atom is given as

$\Delta K_{avg}=\frac{\Delta E_{int}}{N}$

Here$N$is the Avogadro number and is given by$6.023\times {10}^{23}$per mole.

We have$2$of an ideal gas.

$\begin{array}{c}N=2\times 6.023\times {10}^3\\ =12.046\times {10}^{23}\end{array}$

Substitute$12.046\times {10}^{23}$ for$N$ into the above equation

$\begin{array}{c}{\text{ΔK}}_{avg}=\frac{374}{12.046\times {10}^{23}}\text{J}\\ =3.11\times {10}^{-22}\text{J}\end{array}$

Therefore the change intheaverage kinetic energy of gas is$3.11\times {10}^{-22}\text{J}$.