Question

Five moles of monatomic ideal gas have initial pressure $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathit{P}\mathit{a}$and initialvolume $\mathbf{2}\mathbf{.}\mathbf{1}{\mathit{m}}^{\mathbf{3}}$. While undergoing an adiabatic expansion, the gas does 1480J of work. What is the final pressure of the gas after the expansion?

Short answer

The final pressure is $p_2=1.48\times {10}^{3}Pa$.

Step-by-step solution

The equations involved in the process

The relation between pressure and volume in an adiabatic process is ${\mathit{p}}_{\mathbf{1}}{\mathit{V}}_{\mathbf{1}}^{\mathbf{\gamma }}\mathbf{=}{\mathit{p}}_{\mathbf{2}}{\mathit{V}}_{\mathbf{2}}^{\mathbf{\gamma }}$.

The work done in an adiabatic process is$\mathit{W}\mathbf{=}\frac{{\mathbf{p}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}\mathbf{-}{\mathbf{p}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}{\mathbf{\gamma }\mathbf{-}\mathbf{1}}$.

Calculate the final pressure using the two equations

For a monoatomic gas,$\mathit{\gamma }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{67}$.

Given that $n=5,P_1=2.5\times {10}^{3}Pa,V_1=2.1m^3$and $W=1480J$.

$$\begin{gathered} W=\left(\frac{p_1{V}_1-p_2{V}_2}{\gamma -1}\right)/ \\ 1480=(5250-p_2{V}_2)/0.67 \\ {p}_2{V}_2=4258.4J \\ \frac{8630.7}{V_2^{1.67}}\times V_2=4258.4 \\ {V}_2^{-0.67}=0.4934 \\ {V}_2=2.87m^3 \end{gathered}$$

Now, $p_2{V}_2=4258.4$

$$\begin{gathered} p_2=4258.4/2.87 \\ {p}_2=1.48\times {10}^{3}Pa \end{gathered}$$

So, the final pressure is $p_2=1.48\times {10}^{3}Pa$.