Given conditions,
No. of moles(n) = 2 moles.
Temperature(T) = 300 K.
V₂ = 10 L, and V₁ = 1 L.
Universal Gas Constant (R) = 8.31 J/moleK.
Using the formula,
Work done = -2.303nRTlog(V₂/V₁) [For Chemistry.]
∴ Work done = -2.303 × 2 × 8.31 × 300 × log(10/1)
∴ Work done = -11482.758 J.
∴ Work done = -11.483 kJ.
Now, Process is Isothermal, therefore, Internal Energy will also be zero.
Thus, Using first Law of thermodynamics (of Chemistry and not of physics)
∴ ΔU = ΔQ + w
∴ ΔQ = - w
∴ ΔQ = - (-11.483)
∴ ΔQ = 11.483 kJ ≈ 11.4 kJ.
Now, ΔQ is the enthaly because heat constant at constant pressure is enthalpy.
Hence, Option (a). is true.
Hope it helps.