Respuesta :
Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:
(1) < (5) < (3) < (4) < (2)
Explanation :
The formula used for isothermally irreversible expansion is :
[tex]w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)[/tex]
where,
w = work done
[tex]p_{ext}[/tex] = external pressure
[tex]V_1[/tex] = initial volume of gas
[tex]V_2[/tex] = final volume of gas
The expression used for work done in reversible isothermal expansion will be,
[tex]w=-nRT\ln (\frac{V_2}{V_1})[/tex]
where,
w = work done = ?
n = number of moles of gas = 1 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = [tex]25^oC=273+25=298K[/tex]
[tex]V_1[/tex] = initial volume of gas
[tex]V_2[/tex] = final volume of gas
First we have to determine the work done for the following process.
(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.
[tex]w=-p_{ext}(V_2-V_1)[/tex]
[tex]w=-(2.5atm)\times (10-1)L[/tex]
[tex]w=-22.5L.atm=-22.5\times 101.3J=-2279.25J[/tex]
(2) A free isothermal expansion from 1 L to 100 L.
[tex]w=-nRT\ln (\frac{V_2}{V_1})[/tex]
[tex]w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})[/tex]
[tex]w=-11409.6J[/tex]
(3) A reversible isothermal expansion from 0.5 L to 4 L.
[tex]w=-nRT\ln (\frac{V_2}{V_1})[/tex]
[tex]w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})[/tex]
[tex]w=-5151.97J[/tex]
(4) A reversible isothermal expansion from 0.5 L to 40 L.
[tex]w=-nRT\ln (\frac{V_2}{V_1})[/tex]
[tex]w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})[/tex]
[tex]w=-10856.8J[/tex]
(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.
[tex]w=-p_{ext}(V_2-V_1)[/tex]
[tex]w=-(0.5atm)\times (100-1)L[/tex]
[tex]w=-49.5L.atm=-49.5\times 101.3J=-5014.35J[/tex]
Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:
(1) < (5) < (3) < (4) < (2)
