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Question: Calculate the work done in joules when \(3\) moles of an ideal gas at \({27^ \circ }C\) expands isot...

Calculate the work done in joules when 33 moles of an ideal gas at 27∘C{27^ \circ }C expands isothermally and reversibly from 10atm10atm to 1atm1atm (1atm=1.013×105N/m)\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 1atm1atm .

Explanation

Solution

Isothermal reversible process is mostly applicable for ideal gases.it is given by the formula: W=−2.303  nRTlog⁡  V2V1W = - 2.303\;nRT\log \;\dfrac{{{V_2}}}{{{V_1}}} . Where, V1={V_1} = initial volume, V2={V_2} = Final volume, n=n = number of moles, R=R = gas constant, T=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  nRTlog⁡  V2V1W = - 2.303\;nRT\log\; \dfrac{{{V_2}}}{{{V_1}}}….(1)(1)
Where, V1={V_1} = initial volume, V2={V_2} = Final volume, n=n = number of moles, R=R = gas constant, T=T = temperature
Now, Boyle’s law is stated as P1V1=P2V2{P_1}{V_1} = {P_2}{V_2}
V1={V_1} = initial volume, V2={V_2} = Final volume, P1={P_1} = initial pressure, P2={P_2} = Final pressure
But, P1P2=V2V1\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{V_2}}}{{{V_1}}}
Substituting this in equation 11 we get,
W=−2.303nRTlog⁡P1P2W = - 2.303nRT\log \dfrac{{{P_{_1}}}}{{{P_2}}}……(2)(2)
Where, P1={P_1} = initial pressure, P2={P_2} = Final pressure, n=n = number of moles, R=R = gas constant, T=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=3n = 3 moles
T=27∘CT = {27^ \circ }C
T=273+27T = 273 + 27
T=300KT = 300K
P1=10atm{P_1} = 10atm
P2=1atm{P_2} = 1atm
Using the formula of work done in equation 22 we get,
W=−2.303nRTlog⁡P1P2W = - 2.303nRT\log \dfrac{{{P_{_1}}}}{{{P_2}}}

Substituting the values we get,
W=−2.303×3×8.314×300×log⁡101W = - 2.303 \times 3 \times 8.314 \times 300 \times \log \dfrac{{10}}{1}
W=5744.14JW = 5744.14J

Now using the ideal gas equation we will find the work done against the pressure of 1atm1atm
Given data:
Pressure Popp=1atm{P_{opp}} = 1atm
Ideal gas equation: PV=nRTPV = nRT
∴V=nRTP\therefore V = \dfrac{{nRT}}{P}
Where,
V=V = volume of gas, P=P = pressure, n=n = number of moles, R=R = gas constant, T=T = temperature
Now work done is given as:
W=Popp(Vf−Vi)W = {P_{opp}}\left( {{V_f} - {V_i}} \right) …………... (3)(3)
Where, POPP={P_{OPP}} = constant pressure at 1atm1atm
Vf={V_f} = final volume, Vi={V_i} = initial volume
Substituting the values of ideal gas in equation 33 we get,
W=Popp(nRTP2−nRTP1)W = {P_{opp}}\left( {\dfrac{{nRT}}{{{P_2}}} - \dfrac{{nRT}}{{{P_1}}}} \right) ………….. (4)(4)
Where P1={P_1} = initial pressure, P2={P_2} = Final pressure.
Given data: n=3n = 3 moles, P1=10atm{P_1} = 10atm , P2=1atm{P_2} = 1atm , R=8.314R = 8.314 T=300KT = 300K
Substituting these values in equation 44 we get,
W=1×3×8.314×300(11−110)W = 1 \times 3 \times 8.314 \times 300\left( {\dfrac{1}{1} - \dfrac{1}{{10}}} \right)
W=7482.6(910)W = 7482.6\left( {\dfrac{9}{{10}}} \right)
W=6734.34JW = 6734.34J

Therefore, 6734.34J6734.34J is the work done if the expansion is against the constant pressure of 1atm1atm.

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.