Question

When a quantity of monatomic ideal gas expands at a constant pressure of $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathit{P}\mathit{a}$, the volume of the gas increases fromrole="math" localid="1664296449974" $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathit{m}}^{\mathbf{3}}$to $\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathit{m}}^{\mathbf{3}}$. What is the change in the internal energy of the gas?

Short answer

The change in the internal energy would be 360J.

Step-by-step solution

Find a new relation of internal energy

At constant pressure, the formula for the ideal gas is $\mathit{p}\mathit{\Delta }\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathit{\Delta }\mathit{T}$.

Where nrepresents the number of moles, Ris the gas constant and has the value of 8.314J/mol.K. The change in temperature is due to the change in volume.

The change in internal energy for any gas is represented by

$$\begin{gathered} \mathit{\Delta }\mathit{U}\mathbf{=}\mathit{n}{\mathit{C}}_{\mathbf{V}}\mathit{\Delta }\mathit{T} \\ \mathit{\Delta }\mathit{U}\mathbf{=}\mathit{n}\mathbf{\left(}\frac{\mathbf{3}}{\mathbf{2}}\mathit{R}\mathbf{\right)}\mathit{\Delta }\mathit{T} \\ \frac{\mathbf{2}}{\mathbf{3}}\mathit{\Delta }\mathit{U}\mathbf{=}\mathit{n}\mathit{R}\mathit{\Delta }\mathit{T} \end{gathered}$$

Clearly, we can write $\frac{2}{3}∆U=p∆U$.

Put the values and calculate the change in internal energy

Given that $p=4\times {10}^{4}Pa,V_2=8\times {10}^{-3}{m}^3$and $V_1=2\times {10}^{-3}{m}^3$.

$$\begin{gathered} \frac{2}{3}\Delta U=p\Delta V \\ \Delta U=\frac{3}{2}p(V_2-V_1) \\ \Delta U=\frac{3}{2}\times 4\times {10}^4\times (8\times {10}^{-3}-2\times {10}^{-3}) \\ \Delta U=360J \end{gathered}$$

So, the change in the internal energy would be 360J.