What is the significance of work done by an ideal gas?

[AdkinsJr]

See my reply

 

[AdkinsJr]

I'm trying to calculate the work done when 50.0 g of tin dissolves in excess acid at 1.00 atm and 298.15 K

Sn(s)+2H^+(aq)-->Sn^{2+}(aq)+H_2(g)

Here is what I've done:

\Delta V=\frac{\Delta n RT}{P}

\Delta n=n_{H_2(g)(after)}-n_{H_2(g)(before)}=2 mol H_2

It seems that the moles of hydrogen gas will apply pressure to the air, so the work will be

w=-P\Delta V

\frac{-P(2 mol H_2)RT}{P}

=-(2 mol H_2)RT

This makes sense in terms of dimensions, but I don't have the answer in my book, so I can't tell if I did this right.