Question

A monatomic ideal gas that is initially at $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathit{P}\mathit{a}$ and has a volume of$\mathbf{0}\mathbf{.}\mathbf{08}{\mathit{m}}^{\mathbf{3}}$ is compressed adiabatically to a volume of $\mathbf{0}\mathbf{.}\mathbf{04}{\mathit{m}}^{\mathbf{3}}$. (a) What is the final pressure? (b) How much work is done by the gas? (c) What is the ratio of the final temperature of the gas to its initial temperature? Is the gas heated or cooled by this compression?

Short answer

1. $p_2=4.77\times {10}^{5}Pa$
   
2. $W=-10586.38J$
   
3. $\frac{T_2}{T_1}=1.59$and gas is heated

Step-by-step solution

The equations involved in the process

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

In an adiabatic process, the relation between the pressure and volume 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$\mathit{W}\mathbf{=}\frac{{\mathbf{p}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}\mathbf{-}{\mathbf{p}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}{\mathbf{\gamma }\mathbf{-}\mathbf{1}}$.

In an adiabatic process, the relation between the temperature and volume is${\mathit{T}}_{\mathbf{1}}{\mathit{V}}_{\mathbf{1}}^{\mathbf{\gamma }\mathbf{-}\mathbf{1}}\mathbf{=}{\mathit{T}}_{\mathbf{2}}{\mathit{V}}_{\mathbf{2}}^{\mathbf{\gamma }\mathbf{-}\mathbf{1}}$.

Calculate the final pressure

Given that $P_1=1.5\times {10}^{5}Pa,V_1=0.08m^3$and $V_2=0.04m^3$.

$$\begin{gathered} p_2=p_1(\frac{V_1}{V_2}{)}^{\gamma } \\ {p}_2=1.5\times {10}^5\times (\frac{0.08}{0.04}{)}^{1.67} \\ {p}_2=4.77\times {10}^{5}Pa \end{gathered}$$

Calculate the work done

$$\begin{gathered} W=\frac{p_1{V}_1-p_2{V}_2}{\gamma -1} \\ W=\frac{1.5\times {10}^5\times 0.08-4.77\times {10}^5\times 0.04}{1.67-1} \\ W=-10586.38J \end{gathered}$$

Calculate the temperature ratio

$$\begin{gathered} T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1} \\ \frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1} \\ \frac{T_2}{T_1}={\left(\frac{0.08}{0.04}\right)}^{1.67-1} \\ \frac{T_2}{T_1}=1.59 \end{gathered}$$

Since, the final temperature is greater than the initial temperature, gas is heated.