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

Hint: As we know that, when there are changes or expansion in the volume of a gas, the change in work done also occurs. So, we will apply the formula of work done in the terms of change in the volumes of the given sample of a gas, i.e.. $dw = pdv$ .

Complete step by step answer:
As per the question:
When a gas expands at constant pressure then for a small change in volume $'dv'$ , then the work done is, $dw = pdv$ .
where, $dw$is the work done of the change in volume.
Now, if the volume changes from ${v_1}\,to\,{v_2}$ at constant pressure $'p'$ , the change in volume is as, $dv = {v_2} - {v_1}$
Then the work done is:
$dw = p({v_2} - {v_1})$
When the work is done by the system against external pressure then $dw = pdv$ ,
$ \Rightarrow w = \smallint _{{v_1}}^{{v_2}}pdv$
So, by concluding the upper equation by putting the value of given volumes of a gas:
$
   \Rightarrow dw = 1.29atm(17.2L - 12.5L) \\
   \Rightarrow dw = 6.063atm.L \\
 $
Now, as we know that-
$\because 1atm = 101325Pa$
And, we also know that: $1Pa = 1J.{m^{ - 3}}$
So, we get:
$ \Rightarrow 1atm = 101325J.{m^{ - 3}}$ .
Now, as per the question, the work done should be in the units of Joule, so:
$\therefore dw = 6.063 \times 101325\dfrac{J}{{{m^3}}} \times {10^{ - 3}}{m^3} = 614.3334J$
Hence, the work done in the terms of Joules is $614.33joules$ .

Note:
When the volume of a gas increases, the gas performs work (so, if no energy is supplied, the temperature of the gas will decrease). When the volume of a gas drops, an external force exerts work on it (so, if the energy is not allowed to escape, the temperature of the gas will increase).