91Ó°ÊÓ

• Change in temperature is$\mathrm{Δ}T=15.0\text{‿é}$
• The number of moles of gas$n=2.00\text{″¾´Ç±ô}$

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

$Q=n{C}_P\mathrm{Δ}T$

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 $\mathrm{Δ}T$is the temperature difference.

The expression for the work done by the gas is given by,

$W=p\mathrm{Δ}V$

Here $W$is the work done by the gas, $p$is the pressure and$\mathrm{Δ}V$ is the change in the volume.

Work done by the gas$W$;

Since the process is a constant-pressure

$W=p\mathrm{Δ}V$

Using ideal gas equation,

$\begin{array}{c}W=nR\mathrm{Δ}T\\ =\left(2\text{mol}\right)\left(8.31\frac{\text{J}}{\text{mol}.\text{K}}\right)\left(15\text{‿é}\right)\\ =249\text{ J}\end{array}$

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

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,

$⇒C_P=\frac{5}{2}R$

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

$C_P=\frac{Q}{n\mathrm{Δ}T}$

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

$Q=n{C}_P\mathrm{Δ}T$

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

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

The change intheinternal energy of the gas$\mathrm{Δ}{E}_{int}$;

Fromthefirst law of thermodynamics,

$\begin{array}{c}Q=\mathrm{Δ}\text{ }{E}_{int}+W\\ \mathrm{Δ}{E}_{int}=Q-W\end{array}$

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

$\begin{array}{c}\mathrm{Δ}{E}_{int}=623\text{ J}−249\text{ J}\\ =374\text{ J}\end{array}$

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

The change intheaverage kinetic energy per atom $\mathrm{Δ}{K}_{avg}$;

The change in average kinetic energy per atom is given as

$\mathrm{Δ}{K}_{avg}=\frac{\mathrm{Δ}{E}_{int}}{N}$

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

We have$2$of an ideal gas.

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

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

$\begin{array}{c}{\text{Δ°­}}_{avg}=\frac{374}{12.046×{10}^{23}}\text{J}\\ =3.11×{10}^{−22}\text{J}\end{array}$

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