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Q67P

Expert-verifiedFound in: Page 581

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

** An ideal monatomic gas initially has a temperature of 330K ****and a pressure of 6.00atm****. It is to expand from volume 500cm ^{3 }**

- Final Pressure when expansion process is isothermal is 2.00atm.
- Work done by the gas when expansion process is isothermal is 333J.
- Final Pressure when expansion process is adiabatic is 0.96atm.
- Work done by the gas when expansion process is adiabatic is 236J.

Initial Temperature;${T}_{i}=330K$

Initial Pressure;${P}_{i}=6.00atm$

Initial Volume;${V}_{i}=500\hspace{0.33em}c{m}^{3}$

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,**

$W=nRT\hspace{0.33em}ln\left(\frac{{v}_{f}}{{v}_{i}}\right)$…… (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_{i }is 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)

$\mathrm{W}={\mathrm{P}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}\hspace{0.33em}\mathrm{ln}\left(\frac{{\mathrm{v}}_{\mathrm{f}}}{{\mathrm{v}}_{\mathrm{i}}}\right)$

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},500cm^{3} for V_{i}, 1500cm^{3 }for V_{f} into the above equation,

$(6.00\times 500)={P}_{f}\times 1500\phantom{\rule{0ex}{0ex}}{P}_{f}=\frac{(6.00\times 500)}{1500}$

$\Rightarrow {P}_{f}=2.00atm$

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 m^{3} from cm^{3}

Initial pressure

$\begin{array}{ll}{P}_{i}=6atm& \phantom{\rule{0ex}{0ex}}\\ =6atm\times \frac{101\times {10}^{5}pa}{1atm}& {}^{\phantom{\rule{0ex}{0ex}}}\\ =6.06\times {{10}^{5}}^{}& \phantom{\rule{0ex}{0ex}}\end{array}$

Initial volume

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

Final volume

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

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

Substitute $6.06\times {10}^{5}pafor{P}_{i},5\times {10}^{-4}{m}^{3}for{V}_{i},15\times {10}^{-4}{m}^{3}for{V}_{f}$ into the above equation,

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

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

The process is adiabatic so, we can say that,

${P}_{i}{V}_{i}^{\gamma}={P}_{f}{V}_{f}^{\gamma}$

$(6.00\times {500}^{1.67})={P}_{f}\times {1500}^{1.67}$

Substitute $6.00atmfor{P}_{i},500c{m}^{3}for{V}_{i},1500c{m}^{3}for{V}_{f},1.67for\gamma $ into the above equation,

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

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

The process is adiabatic so, we can say that,

$W={P}_{i}{V}_{i}^{\gamma}{\int}_{{V}_{i}}^{{V}_{f}}{V}^{-\gamma}dV$

On integrating the above equation,

localid="1662538514457" $\begin{array}{l}w={P}_{i}{V}_{i}\times \left(\frac{{V}_{f}^{1-\gamma}}{1-\gamma}\right)\\ \\ =\frac{{P}_{f}{V}_{f}-{P}_{i}{V}_{i}}{1-\gamma}\end{array}$

Convert the initial and final pressure from atm to pa

Initial pressure

localid="1662539084992" ${P}_{i}=6atm\phantom{\rule{0ex}{0ex}}=6atm\times \frac{101325pa}{1atm}\phantom{\rule{0ex}{0ex}}=607950pa$${P}_{f}=0.96atm\phantom{\rule{0ex}{0ex}}=0.96atm\times \frac{101325pa}{1atm}\phantom{\rule{0ex}{0ex}}=97272pa$

Final pressure

${P}_{f}=0.96atm\phantom{\rule{0ex}{0ex}}=0.96atm\times \frac{101325pa}{1atm}\phantom{\rule{0ex}{0ex}}=97272pa$

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

$W=({P}_{f}{V}_{f}-{P}_{i}{V}_{i})/(1-\gamma )$

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

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

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

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