91Ó°ÊÓ

• Number of moles of an ideal monoatomic gas,n=2.00
• Temperature difference,$\mathrm{Δ}T=15.0K$
• Change in volume,$\mathrm{Δ}V=0$

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 volume 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 is;

$W=p\mathrm{Δ}V$

But,

$\mathrm{Δ}V=0$

Hence,

W=0

Therefore, the work W done by the gas is zero.

Heat energy is given by

$Q=n{C}_v\mathrm{Δ}T$$Q=n{C}_v\mathrm{Δ}T$

But, for a monoatomic gas,

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

Substitute role="math" localid="1662468781681" $\frac{3}{2}R$for C_V into the above equation,

$Q=n\left(\frac{3}{2}R\right)\mathrm{Δ}T$
role="math" localid="1662469017230" $$\begin{gathered} =2\left(\frac{3}{2}8.314\right)\left(15.0\right) \\ =374.13 \\ =374J \end{gathered}$$

Therefore, the energy transferred as heat Q is 374J.

According to the first law of thermodynamics,

$\mathrm{Δ}{E}_{int}=Q-W$

Substitute 374J for Q and 0J for W into the above equation,

role="math" localid="1662469480888" $$\begin{gathered} \mathrm{Δ}{E}_{int}=374-0 \\ =374J \end{gathered}$$

Therefore, the change $\mathrm{Δ}{E}_{int}$in the internal energy of the gas is 374J.

No. of atoms (N) = No. of moles x Avogadro no.

$\begin{array}{l}N=2×6.02×{10}^{23}\\ =12.04×{10}^{23}\end{array}$

$Changeintheaveragekineticenergyperatom\left(\mathrm{Δ}K\right)=\frac{\left(\mathrm{Δ}{E}_{i}nt\right)}{N}$

Substitute 374J for$\mathrm{Δ}{E}_{i}nt$and $12.04×{10}^{23}$for into the above equation,

role="math" localid="1662470169132" $\begin{array}{l}K=\frac{374}{12.04×{10}^{23}}\\ =31.063×{10}^{-23}\\ =3.11×{10}^{-22}\end{array}$

Therefore, the change in the average kinetic energy per atom is$3.11×{10}^{-22}J$