Answer : The work done on the gas will be, 418.4 J
Explanation :
First we have to calculate the volume at 270°C.

where,
P = pressure of gas = 1 atm
= volume of gas = ?
T = temperature of gas = 
n = number of moles of gas = 0.205 mol
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:


Now we have to calculate the volume at 24°C.

where,
P = pressure of gas = 1 atm
= volume of gas = ?
T = temperature of gas = 
n = number of moles of gas = 0.205 mol
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:


Now we have to calculate the work done.
Formula used :

where,
w = work done
p = pressure of the gas = 1 atm
= initial volume = 9.12 L
= final volume = 4.99 L
Now put all the given values in the above formula, we get:



conversion used : (1 L.atm = 101.3 J)
Therefore, the work done on the gas will be, 418.4 J