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 ((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)* (10-1)L`

`w=-22.5L.atm=-22.5* 101.3J=-2279.25J`

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

`w=-nRT\ln ((V_2)/(V_1))`

`w=-1mole* 8.314J/moleK* 298K* \ln ((100L)/(1L))`

`w=-11409.6J`

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

`w=-nRT\ln ((V_2)/(V_1))`

`w=-1mole* 8.314J/moleK* 298K* \ln ((4L)/(0.5L))`

`w=-5151.97J`

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

`w=-nRT\ln ((V_2)/(V_1))`

`w=-1mole* 8.314J/moleK* 298K* \ln ((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)* (100-1)L`

`w=-49.5L.atm=-49.5* 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)