Question

Three moles of an ideal monatomic gas expands at a constant pressure of 2.5atm ; the volume of the gas changes from $\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathit{m}}^{\mathbf{3}}$to $\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathit{m}}^{\mathbf{3}}$. Calculate (a) the initial and final temperatures of the gas; (b) the amount of work the gas does in expanding; (c) the amount of heat added to the gas; (d) the change in internal energy of the gas.

Short answer

$$\begin{gathered} a.T_1=325K \\ {T}_2=457K \\ b.W=3293J \\ c.Q=8231J\Delta \\ d.U=4938J \end{gathered}$$

Step-by-step solution

The equations involved

The ideal gas equation is pV = nRT.

Work done at constant pressure is given by$\mathit{W}\mathbf{=}\mathit{p}\mathbf{∆}\mathit{V}$.

The heat energy is given by$\mathit{Q}\mathbf{=}\mathit{n}{\mathit{C}}_{\mathbf{P}}\mathit{\Delta }\mathit{T}$.

First law of thermodynamics is$\mathit{Q}\mathbf{=}\mathit{W}\mathbf{+}\mathit{\Delta }\mathit{U}$.

Use ideal gas equation to find the temperatures

$$\begin{gathered} T_1=\frac{p{V}_1}{nR} \\ {T}_1=\frac{2.5\times 101325\times 3.2\times {10}^{-2}}{3\times 8.314} \\ {T}_1=325K \\ {T}_2=\frac{p{V}_2}{nR} \\ {T}_2=\frac{2.5\times 101325\times 4.5\times {10}^{-2}}{3\times 8.314} \\ {T}_2=457K \end{gathered}$$

Calculate the work done

$$\begin{gathered} W=p(V_2-V_1) \\ W=2.5\times 101325\times (0.045-0.032) \\ W=3293J \end{gathered}$$

Calculate the heat energy involved

$$\begin{gathered} Q=n{C}_P\Delta T \\ Q=n\times 2.5R\Delta T \\ Q=3\times 2.5\times 8.314\times (457-325) \\ Q=8231J \end{gathered}$$

Calculate the internal energy

$$\begin{gathered} Q=W+\Delta U \\ \Delta U=Q-W \\ \Delta U=8231-3293 \\ \Delta U=4938J \end{gathered}$$