Question

One mole of an ideal monatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, $27^\circ C$. The work done on the gas will be
(A) $300R\ln 6$
(B) $300R$
(C) $300R\ln 7$
(D) $300R\ln 2$

Answer 1

Hint: The gas is compressed isothermally at room temperature. So, the temperature will remain constant throughout the process. As you know, that the work done is equal to the integral of force over displacement, find the work done in terms of pressure and change in volume. If necessary, use the ideal gas equation to find unknown quantities such as volume.

Complete step by step answer:
Consider a cylinder containing one mole of an ideal monatomic gas with a movable piston having area \[A\]. In order to increase the pressure of the gas, you apply force on the piston so that the piston moves down and the pressure will increase. Let the force applied by you be $F$. If the piston moves by a distance of $dx$, then the work done by you or the work done on the system will be $dW = Fdx$. Now, force per unit area is defined as pressure, therefore, pressure exerted by you on the piston is ${P_{ext}} = \dfrac{F}{A} \to F = {P_{ext}}A$. Substituting this value of force in the expression for work, we get, $dW = {P_{ext}}Adx = {P_{ext}}\left( {Adx} \right)$. As you can see that $Adx = - dV$, that is change in the volume of the gas. The negative sign is because the volume decreases as you exert pressure on the piston. Here, the whole process is done very slowly, so that the pressure exerted by you is actually equal to the pressure of the gas. Hence, we write ${P_{ext}} = P$, where $P$ is the pressure of the gas.
So, we have $dW = - PdV$. The net work done will be the integral of $dW$.
$W = \int {dW = - \int\limits_{{V_1}}^{{V_2}} {PdV} } $. From the ideal gas equation, we can obtain the pressure of the gas,
$\implies PV = nRT \to P = \dfrac{{nRT}}{V}$
$\implies W = - \int\limits_{{V_1}}^{{V_2}} {\dfrac{{nRT}}{V}dV} $, since, it is an isothermal process, $T$ is constant.
$\implies W = - nRT\int\limits_{{V_1}}^{{V_2}} {\dfrac{{dV}}{V}} = - nRT\ln V_{{V_1}}^{{V_2}} = - nRT\left( {\ln {V_2} - \ln {V_1}} \right) = - nRT\ln \dfrac{{{V_2}}}{{{V_1}}}$.
So, the work done in an isothermal reversible process is given as$ - nRT\ln \dfrac{{{V_2}}}{{{V_1}}} = nRT\ln \dfrac{{{V_1}}}{{{V_2}}}$.
From the given data in the question, we have $n = 1$ and $T = 27^\circ C = \left( {273 + 27} \right)K = 300K$.
$W = 300R\ln \dfrac{{{V_1}}}{{{V_2}}}$
Now, let us use the ideal gas equation,
$
 PV = nRT \\
\implies {P_1}{V_1} = nRT \\
\implies {P_2}{V_2} = nRT \\
 \implies {P_1}{V_1} = {P_2}{V_2} \\
 \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{{V_1}}}{{{V_2}}} \\
 $
As the pressure is doubled, we have ${P_2} = 2{P_1}$
$
 \dfrac{{2{P_1}}}{{{P_1}}} = \dfrac{{{V_1}}}{{{V_2}}} \\
\implies \dfrac{{{V_1}}}{{{V_2}}} = 2 \\
 $
Substituting this value of ratio of initial volume and final volume in the expression of work, we get, $W = 300R\ln 2$.
Therefore, the work done on the gas will be $300R\ln 2$.

So, the correct answer is “Option D”.

Note:
Remember that the process is reversible and the work done in an isothermal reversible process is equal to $nRT\ln \dfrac{{{V_1}}}{{{V_2}}}$. Also keep in mind that the temperature is always taken in kelvins, so do not forget to convert the given temperature in kelvins. Look at that method carefully and understand how the work was calculated using force and then substituting it with appropriate quantities, that are pressure and area. The whole process is carried out slowly in order to maintain the equilibrium, that is external pressure equals the pressure of the gas at any instant of time.