question on solution for 8.49

[Lindsay H 2B]

8.49: Oxygen difluoride reacts rapidly with water vapor to produce O2, HF and heat:
OF2(g) + H2O(g) --> O2(g) + 2HF(g) ∆H= -318 kJ
What is the change in internal energy for the reaction of 1.00 mol OF2?

The solutions manual shows calculating the work portion of the change in internal energy by using P∆V=∆nRT, and seems to assume that T=298K. Can anybody explain how we would get this value of T?

[Aijun Zhang 1D]

If nothing is specified, the standard temperature for gas reaction is 298K.

[zanekoch1A]

Would another way to solve it be knowing that for a equation like this it is appropriate to assume it is at constant pressure which would mean change in internal nrg= q which is in turn equal to, and constant pressure, enthalpy change which is given? The answer would then be -318, the book gives -320 but is this just because of sig figs?