1.02 L
W = -P(V2 - V1)
where
W = Work
P = Pressure
V1 = Starting volume
V2 = Ending volume
Since the gas performed the work, the work needs to be negatively signed, so W = -235J. Since we have mixed units of atm and joules, we need to convert the pressure from atm to something more convenient. We can use a couple of the varients of the universal gas constant with different units. I'll use
R = 8.314472 J/(mol*K) and R = 0.082057 L*atm/(mol*K), dividing the second by the first. So
0.082057 L*atm/(mol*K) / 8.314472 J/(mol*K) = 0.009869178 L*atm/J will be our conversion factor.
Now for the math
V2 = -W/P + V1
V2 = -(-235J * 0.009869178 L*atm/J) / 3.00 atm + 0.250 L
V2 = -(-2.319256713 L*atm) / 3.00 atm + 0.250 L
V2 = -(-0.773085571 L) + 0.250 L
V2 = 0.773085571 L + 0.250 L
V2 = 1.023085571 L
Rounding to 3 significant figures gives 1.02 L
