Question

One mole of oxygen of volume $1litre$at $4atm$pressure to attain $1atm$pressure by the result of isothermal expansion. Find work done by the gas.
A. $155J$
B. $206J$
C. $355J$
D. $552J$

Answer 1

Hint
In the isothermal process, the temperature remains constant. So during this isothermal expansion, the initial temperature is equal to the final temperature. To find the temperature, use the equation of ideal gas law and then find work done by the gas using its formula.

Formulae used:
$PV = nRT$; $W = 2.303RT\log \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right)$

Complete step-by-step solution :So, a volume of gas is expanded by the work done through an isothermal process.
We have, work done in isothermal expansion = $W = 2.303RT\log \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right)$.
Here, $R =$universal gas constant $= 0.0821atm/mol/K$
$T =$ Temperature throughout the process of Isothermal expansion, since the temperature is constant
${P_1} =$ Initial pressure (before expansion)
${P_2} =$ Final pressure (after expansion)
Now, to find the temperature, we apply the ideal gas law for initial conditions.
Ideal gas law: $PV = nRT$
Where, $n$= mole of gas
Initial conditions:
${P_1} = 4atm$
${V_1} = 1litre$
$n = 1mole$
And, $T = {T_1} = {T_2} = ?$
By substituting these values in ideal gas law, we get
$nRT = PV = 4 \times 1 \\\ \Rightarrow nRT = 4 \\\$
Now, work done in isothermal expansion is
$W = 2.303RT\log \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right)$
Since, $RT = \dfrac{{RT}}{n} = 4$
And, $\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{4}{1} = 4$
So, by putting these values in the equation of $W$, we get
$W = 2.303 \times 4 \times \log \left( 4 \right)$
By further simplifying this, we have
$W = 2.303 \times 4 \times .6 \\\ W = 5.5272L - atm \\\$
As units of work done that we got is $L - atm$ but the given options are in joule. So we have to convert the units.
For that, we have
$1L - atm \approx 100J$
So, work done becomes
$W \approx 5.5272 \times 100J \\\ W \approx 552.72J \\\$
Since the nearest option is D, that is, $W = 552J$, therefore the correct answer is option D.

Note:-
$\left. 1 \right)$ The temperature of the isothermal process is constant because the transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained. And this all done if a system is in contact with a thermal reservoir from outside, then, to maintain the thermal equilibrium, the system adjusts itself slowly with the reservoir’s temperature through heat exchange.
$\left. 2 \right)$ Here the work is done by the system on surrounding.