Question

What is the internal energy of$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{o}\mathbf{\text{l}}$of an ideal monoatomic gas at$\mathbf{273}\mathit{K}$?

Short answer

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

Step-by-step solution

Given data

• Temperature of ideal monoatomic gas,$\text{T}=273\text{\hspace{0.17em}K}$
• Number of moles,$\text{n}=1.0$

Understanding the concept

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

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

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

Calculate the internal energy of  1.0  molof an ideal monoatomic gas at

For an ideal monoatomic gas,

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

$C_V=\frac{\Delta E_{int}}{n\Delta T}$...... (2)

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

$\begin{array}{c}\Delta E_{int}=n{C}_V\Delta T\\ =\frac{3}{2}nRT\end{array}$

Here R is the gas constant; its value is$R=8.31\text{\hspace{0.17em}J}/\text{mol.K}$

Substituting all values in the above equation, we get

$\begin{array}{c}\Delta {\text{E}}_{int}=\frac{3}{2}\left(1.0\text{\hspace{0.17em}mol}\right)(8.31\text{\hspace{0.17em}J}/\text{mol.K})\left(273\text{\hspace{0.17em}K}\right)\\ =3.4\times {10}^3\text{J}\end{array}$

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