Calculate the amount of work done when one mole of an ideal gas contained in a bulb of `2 L` capacity at 1 atm is allowed to enter in evacuated bulb of 10 L capacity and in another case it is allowed to expand from 1 L to 5 L against a pressure of 1 atm.

Explanation

(i) `W=-P_(ext)xx Delta V`
As expansion takes place into the evacuated bulb, i.e., against vacuum, `P_(ext)=0`. Hence W = 0.
(ii) `Delta V=V_(2)-V_(1)=5-1=4` litres, P = 1 atm
`therefore W = -P Delta V = -1xx4` litre atm `=-4xx101.3J=-405.2 J`
Alternatively, using the SI units directly
P = 1 atm = 101325 Pa, `Delta V=4L=4xx10^(-3)m^(3)`
`therefore W =-P xx Delta V=-101325xx4xx10^(-3)J=-405.3 J`
The negative sign implies that the work is done by the system.