Question

One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure. The change in an internal energy of the gas during the transition is?

Answer 1

Step 1: Internal energy

The formula to find the internal energy is$\mathrm{\Delta U}={\mathrm{nC}}_{\mathrm{v}}\mathrm{dT}.....\left(1\right)$

Where $\mathrm{n}$ is no. of moles, ${\mathrm{C}}_{\mathrm{v}}$ is heat capacity at constant volume, and $\mathrm{dT}$ is temperature change (${\mathrm{T}}_2-{\mathrm{T}}_1$)

For diatomic gas,${\mathrm{C}}_{\mathrm{v}}=\frac{5}{2}\mathrm{R}$

From the ideal gas equation,${\mathrm{T}}_1=\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{\mathrm{R}}$and $T_2=\frac{P_2{V}_2}{R}$

Substituting these in equation $\left(1\right)$ we get,

$$\begin{gathered} \mathrm{\Delta U}={\mathrm{C}}_{\mathrm{v}}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right) \\ \mathrm{\Delta U}=\frac{5}{2}\mathrm{R}\left(\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{\mathrm{R}}-\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{\mathrm{R}}\right) \\ \mathrm{\Delta U}=\frac{5}{2}\left({\mathrm{P}}_2{\mathrm{V}}_2-{\mathrm{P}}_1{\mathrm{V}}_1\right)......\left(2\right) \end{gathered}$$

Step 2: Substituting the values

From the graph,

$$\begin{gathered} {\mathrm{P}}_1=5\mathrm{kPa} \\ {\mathrm{P}}_2=2\mathrm{kPa} \\ {\mathrm{V}}_1=4{\mathrm{m}}^3 \\ {\mathrm{V}}_2=6{\mathrm{m}}^3 \end{gathered}$$

Therefore, by substituting these values in the equation $\left(2\right)$ we get,

$$\begin{gathered} \mathrm{\Delta U}=\frac{5}{2}\left({\mathrm{P}}_2{\mathrm{V}}_2-{\mathrm{P}}_1{\mathrm{V}}_1\right) \\ \mathrm{\Delta U}=\frac{5}{2}\left(2\times 6-5\times 4\right) \\ \mathrm{\Delta U}=\frac{5}{2}\left(12-20\right) \\ \mathrm{\Delta U}=\frac{5}{2}\times -8 \\ \mathrm{\Delta U}=5\times -4 \\ \mathrm{\Delta U}=-20\mathrm{kJ} \end{gathered}$$

Therefore, the change in the internal energy of the gas during the transition is $-20\mathrm{kJ}$.