Final answer:
To find the work done during the expansion of H₂, we calculate the work for each stage of expansion separately and then sum them to get a total of approximately -727 cal.
Step-by-step explanation:
To determine the work (w) done when 2.6 moles of H₂ gas expands first from 25 L to 50 L against an external pressure of 1 atm, and then from 50 L to 100 L against an external pressure of 0.1 atm, we use the formula for pressure-volume work: w=-P₎ₑ∆V, where P₎ₑₑ is the external pressure and ∆V is the change in volume.
The work for the first expansion: w=-P₎ₑₑ(V₂-V₁) where P₎ₑₑ = 1 atm, V₂ = 50 L, V₁ = 25 L, thus w = -(1 atm)(50 L - 25 L) = -25 L·atm.
Converting to cal: -25 L·atm x (101.3 J/L·atm) x (1 cal/4.184 J) = -606 cal (approx.)
The work for the second expansion: w = -(0.1 atm)(100 L - 50 L) = -5 L·atm.
Converting to cal: -5 L·atm x (101.3 J/L·atm) x (1 cal/4.184 J) = -121 cal (approx.)
The total work done is the sum of both expansions: -606 cal + (-121 cal) = -727 cal (approx.)