91Ó°ÊÓ

Initial Temperature;$T_i=330K$

Initial Pressure;$P_i=6.00atm$

Initial Volume;$V_i=500 c{m}^3$

Final Volume;$V_f=1500c{m}^3$

In case of isothermal process temperature remains constant. The expression for the work done in case of isothermal process is given by,

$W=nRT ln\left(\frac{v_f}{v_i}\right)$…… (i)

Here W is the work done, n is the number of moles, R is the gas constant, T is the temperature, v_f is the final volume of the gas, v_iis the initial volume of the gas.

From the ideal gas equation;

PV=nRT.........(ii)

Here P is the pressure and V is the volume.

From equation (i) and (ii)

$\mathrm{W}={\mathrm{P}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}} \ln \left(\frac{{\mathrm{v}}_{\mathrm{f}}}{{\mathrm{v}}_{\mathrm{i}}}\right)$

The process is isothermal, we can say that,$P_i{V}_i=P_f{V}_f$

$P_i{V}_i=P_f{V}_f$

Substitute 6.00atm for P_i,500cm³ for V_i, 1500cm³for V_f into the above equation,

$$\begin{gathered} (6.00×500)=P_f×1500 \\ {P}_f=\frac{(6.00×500)}{1500} \end{gathered}$$

$⇒P_f=2.00atm$

Therefore the final pressure when the expansion is isothermal is 2.00atm.

The process is isothermal, we can say that

$W=P_i{V}_{i}ln\left(\frac{v_f}{v_i}\right)$

We have to convert the pressure to Pascal from and Volume to m³ from cm³

Initial pressure

$$\begin{gathered} \begin{array}{ll}{P}_i=6atm& \\ =6atm×\frac{101×{10}^{5}pa}{1atm}& {}^{ \\ }\\ =6.06×{{10}^5}^{}& \end{array} \end{gathered}$$

Initial volume

$\begin{array}{l}{v}_i=500{\mathrm{cm}}^3\\ =500{\mathrm{cm}}^3×\frac{1×{10}^{-6}{m}^3}{1{\mathrm{cm}}^3}\\ =5×{10}^{-4}{m}^3\end{array}$

Final volume

$\begin{array}{l}{v}_t=1500{\mathrm{cm}}^3\\ =1500{\mathrm{cm}}^3×\frac{{1×{10}^{-6}m}^3}{1{\mathrm{cm}}^3}\\ =15×{10}^{-4}{m}^3\end{array}$

The expression for the work done in case of isothermal process is given by,

Substitute $6.06×{10}^{5}pafor{P}_i,5×{10}^{-4}{m}^{3}for{V}_i,15×{10}^{-4}{m}^{3}for{V}_f$into the above equation,

$\begin{array}{l}w=6.06x{10}^5×\left(5×{10}^{-4}\right)×In\left(\frac{15×{10}^{-4}}{5×{10}^{-4}}\right)\\ \\ =332.87\\ =333J\end{array}$

Therefore the work done by the gas when the expansion is isothermal is 333J.

The process is adiabatic so, we can say that,

$P_i{V}_i^{\mathrm{γ}}=P_f{V}_f^{\mathrm{γ}}$

$(6.00×{500}^{1.67})=P_f×{1500}^{1.67}$

Substitute $6.00atmfor{P}_i,500c{m}^{3}for{V}_i,1500c{m}^{3}for{V}_f,1.67for\mathrm{γ}$into the above equation,

$\begin{array}{l}{P}_f=\frac{(6.00×{500}^{1.67})}{{1500}^{1.67}}\\ =0.957\\ =0.96atm\end{array}$

The final pressure when the expansion is adiabatic is 0.96atm.

The process is adiabatic so, we can say that,

$W=P_i{V}_i^{\mathrm{γ}}{∫}_{V_i}^{V_f}{V}^{-\mathrm{γ}}dV$

On integrating the above equation,

localid="1662538514457" $\begin{array}{l}w=P_i{V}_i×\left(\frac{V_f^{1-\mathrm{γ}}}{1-\mathrm{γ}}\right)\\ \\ =\frac{P_f{V}_f-P_i{V}_i}{1-\mathrm{γ}}\end{array}$

Convert the initial and final pressure from atm to pa

Initial pressure

localid="1662539084992" $$\begin{gathered} P_i=6atm \\ =6atm×\frac{101325pa}{1atm} \\ =607950pa \end{gathered}$$$$\begin{gathered} P_f=0.96atm \\ =0.96atm×\frac{101325pa}{1atm} \\ =97272pa \end{gathered}$$

Final pressure

$$\begin{gathered} P_f=0.96atm \\ =0.96atm×\frac{101325pa}{1atm} \\ =97272pa \end{gathered}$$

The expression for work done by the gas in case of adiabatic process is calculated above;

$W=(P_f{V}_f-P_i{V}_i)/(1-\mathrm{γ})$

Substitute $97272pafor{P}_f,607950pafor{P}_i,5×{10}^{-4}{m}^{3}for{V}_{i}and15×{10}^{-4}{m}^{3}for{V}_f$into the above equation,

$\begin{array}{l}w=\frac{\left(97272×15×{10}^{-4}\right)-\left(607950×5×{10}^{-4}\right)}{1-1.67}\\ =235.9\\ =236J\end{array}$

Therefore the work done by the gas when the expansion is adiabatic is 236J.