If your complete question is:

One mole of an ideal gas at 300 K in thermal contact with the surroundings expands isothermally from1 L to2L against a constant pressure of 3 atm. In this process the change in the enthalapy of surroundings is? (change in S in J /K )1Latm.=101.3J

Ans: From First Law of Thermodynamics,

q(system)=del U- del W

Since its an isothermal reaction, del U = 0, del W = -P(ext.) x del V

so q(system) = 0-(-P(ext) x del V)

= 3 x (2-1)

= 3 L.atm

change in enthalphy of system = del S(system) = q/T = 3L.atm/300k = (3 x 101.3)J/300k

= 1.013 J/K

So, change in enthalphy of surrounding = del S(surrounding) = - del S(system) =

- 1.013 J/K