Question

What is the work done when a sample of gas expands from 12.5L to 17.2L against a pressure of 1.29 atm? Give your answer in Joules.

Answer 1

The work done is 614.33 Joules.

Explanation:

When a gas expands at constant pressure then for a small change in volume `'dv'` workdone is `dw=pdv`.
If the volume changes from `'v_1'` to `'v_2'` at constant pressure `'p'`, the work done is `dw=p(v_2-v_1)`.
When work is done by the system against external pressure then `dw=pdv`
`rArrw=int_(v_1)^(v_2) p dv`
So, by calculating with given numericals we get,
`rArr dw=1.29atm(17.2L-12.5L)=1.29atmxx(4.7L)=6.063atm.L`.

To have the answer in terms of we use below units:
`1 atm = 101325 Pa`

But`1 Pa = 1 J//m^3` `(`because `J=Pa.m^3` `)`.

Substitute '`1Pa`' value to get `'1atm`' value,
`rArr 1atm=101325J//m^3`.
Now as we know `1L`itre=`10^-3m^3`.

Converting the workdone `'dw'` value into Joules using the above required units we get,
`dw=6.063xx101325J/cancel(m^3)xx10^-3cancel(m^3)=614.3334J`.
`:.` work done in terms of Joules is `614.33J`oules.