Question

Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 25°C. 1. An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm. 2. A free isothermal expansion from 1 L to 100 L. 3. A reversible isothermal expansion from 0.5 L to 4 L. 4. A reversible isothermal expansion from 0.5 L to 40 L. 5. An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

Answer 1

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 :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

The expression used for work done in reversible isothermal expansion will be,

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $25^oC=273+25=298K$

$V_1$ = initial volume of gas

$V_2$ = 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.

$w=-p_{ext}(V_2-V_1)$

$w=-(2.5atm)\times (10-1)L$

$w=-22.5L.atm=-22.5\times 101.3J=-2279.25J$

(2) A free isothermal expansion from 1 L to 100 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})$

$w=-11409.6J$

(3) A reversible isothermal expansion from 0.5 L to 4 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})$

$w=-5151.97J$

(4) A reversible isothermal expansion from 0.5 L to 40 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})$

$w=-10856.8J$

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(0.5atm)\times (100-1)L$

$w=-49.5L.atm=-49.5\times 101.3J=-5014.35J$

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)