91Ó°ÊÓ

• Temperature of ideal monoatomic gas,$\text{T}=273\text{‿é}$
• Number of moles,$\text{n}=1.0$

The molar specific heat$C_V$of gas at constant volume is given as

$\begin{array}{c}{C}_V=\frac{Q}{n\mathrm{Δ}T}\\ =\frac{\mathrm{Δ}{E}_{int}}{n\mathrm{Δ}T}\end{array}$

Here Qis the energy transferred as heat to or from a sample of nmole of the gas,$\mathrm{Δ}T$is the resulting change in temperature of the gas, and$\mathrm{Δ}{E}_{int}$is the change in internal energy of the gas.

For an ideal monoatomic gas,

$C_V=\frac{3}{2}R$...... (1)

$C_V=\frac{\mathrm{Δ}{E}_{int}}{n\mathrm{Δ}T}$...... (2)

Substituting (1) in (2) and solving for internal energy, we get

$\begin{array}{c}\mathrm{Δ}{E}_{int}=n{C}_V\mathrm{Δ}T\\ =\frac{3}{2}nRT\end{array}$

Here R is the gas constant; its value is$R=8.31\text{ J}/\text{mol.K}$

Substituting all values in the above equation, we get

$\begin{array}{c}\mathrm{Δ}{\text{E}}_{int}=\frac{3}{2}\left(1.0\text{″¾´Ç±ô}\right)(8.31\text{ J}/\text{mol.K})\left(273\text{‿é}\right)\\ =3.4×{10}^3\text{J}\end{array}$

Therefore the internal energy of $1.0\text{ }mol$of an ideal monoatomic gas at$273\text{ }K$ is.$3.4×{10}^3\text{J}$