Question

Calculate the work done in joules when $3$ moles of an ideal gas at ${27^ \circ }C$ expands isothermally and reversibly from $10atm$ to $1atm$ $\left( {1atm = 1.013 \times {{10}^5}N/m} \right)$ . What will be the work done if the expansion is against the constant pressure of $1atm$ .

Answer 1

Hint: Isothermal reversible process is mostly applicable for ideal gases.it is given by the formula: $W = - 2.303\;nRT\log \;\dfrac{{{V_2}}}{{{V_1}}}$ . Where, ${V_1} = $ initial volume, ${V_2} = $ Final volume, $n = $ number of moles, $R = $ gas constant, $T = $ temperature

This process takes place under constant temperature.


Complete step by step answer:
Isothermal reversible process is the process in which the temperature remains constant throughout the reaction only the pressure and volume changes.
isothermal reversible process is given by the formula,
$W = - 2.303\;nRT\log\; \dfrac{{{V_2}}}{{{V_1}}}$….$(1)$
Where, ${V_1} = $ initial volume, ${V_2} = $ Final volume, $n = $ number of moles, $R = $ gas constant, $T = $ temperature
Now, Boyle’s law is stated as ${P_1}{V_1} = {P_2}{V_2}$
${V_1} = $ initial volume, ${V_2} = $ Final volume, ${P_1} = $ initial pressure, ${P_2} = $ Final pressure
But, $\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{V_2}}}{{{V_1}}}$
Substituting this in equation $1$ we get,
$W = - 2.303nRT\log \dfrac{{{P_{_1}}}}{{{P_2}}}$……$(2)$
Where, ${P_1} = $ initial pressure, ${P_2} = $ Final pressure, $n = $ number of moles, $R = $ gas constant, $T = $ temperature
Therefore Boyle’s law is applicable for isothermal reversible processes.
We will use this formula to calculate the work done.

Given data:
$n = 3$ moles
$T = {27^ \circ }C$
$T = 273 + 27$
$T = 300K$
${P_1} = 10atm$
${P_2} = 1atm$
Using the formula of work done in equation $2$ we get,
$W = - 2.303nRT\log \dfrac{{{P_{_1}}}}{{{P_2}}}$

Substituting the values we get,
$W = - 2.303 \times 3 \times 8.314 \times 300 \times \log \dfrac{{10}}{1}$
$W = 5744.14J$

Now using the ideal gas equation we will find the work done against the pressure of $1atm$
Given data:
Pressure ${P_{opp}} = 1atm$
Ideal gas equation: $PV = nRT$
$\therefore V = \dfrac{{nRT}}{P}$
Where,
$V = $ volume of gas, $P = $ pressure, $n = $ number of moles, $R = $ gas constant, $T = $ temperature
Now work done is given as:
$W = {P_{opp}}\left( {{V_f} - {V_i}} \right)$ …………... $(3)$
Where, ${P_{OPP}} = $ constant pressure at $1atm$
${V_f} = $ final volume, ${V_i} = $ initial volume
Substituting the values of ideal gas in equation $3$ we get,
$W = {P_{opp}}\left( {\dfrac{{nRT}}{{{P_2}}} - \dfrac{{nRT}}{{{P_1}}}} \right)$ ………….. $(4)$
Where ${P_1} = $ initial pressure, ${P_2} = $ Final pressure.
Given data: $n = 3$ moles, ${P_1} = 10atm$ , ${P_2} = 1atm$ , $R = 8.314$ $T = 300K$
Substituting these values in equation $4$ we get,
$W = 1 \times 3 \times 8.314 \times 300\left( {\dfrac{1}{1} - \dfrac{1}{{10}}} \right)$
$W = 7482.6\left( {\dfrac{9}{{10}}} \right)$
$W = 6734.34J$

Therefore, $6734.34J$ is the work done if the expansion is against the constant pressure of $1atm$.

Note: In an isothermal reversible process the reaction takes place very slowly and is at equilibrium because of which it fails to tell the time that is taken for the reaction to be completed. - Work done depends on the conditions in which the reaction is being carried out and not on the initial and final state of the reaction.