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### Fundamentals Of Physics

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 500cm3 to volume1500cm3. If the expansion is isothermal, what are (a) the final pressure and (b) the work done by the gas? If, instead, the expansion is adiabatic, what are (c) the final pressure and (d) the work done by the gas?

1. Final Pressure when expansion process is isothermal is 2.00atm.
2. Work done by the gas when expansion process is isothermal is 333J.
3. Final Pressure when expansion process is adiabatic is 0.96atm.
4. Work done by the gas when expansion process is adiabatic is 236J.
See the step by step solution

## Step 1: Write the given data from the question:

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

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

Initial Volume;${V}_{i}=500 c{m}^{3}$

Final Volume; ${V}_{f}=1500c{m}^{3}$

## Step 2: Understanding the concept

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, vf is the final volume of the gas, vi 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)

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

## Step 3: (a) Calculate the final pressure when the expansion is isothermal

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 Pi,500cm3 for Vi, 1500cm3 for Vf into the above equation,

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

$⇒{P}_{f}=2.00atm$

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

## Step 4: (b) Calculate the work done by the gas when the expansion is isothermal

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

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

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.

## Step 5: (c) Calculate the final pressure when the expansion is adiabatic

The process is adiabatic so, we can say that,

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

$\left(6.00×{500}^{1.67}\right)={P}_{f}×{1500}^{1.67}$

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

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

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

## Step 6: (d) Calculate the work done by the gas when the expansion is adiabatic

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

Convert the initial and final pressure from atm to pa

Initial pressure

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

Final pressure

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

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

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

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.

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