# Work done by an Ideal Gas

1. Jan 18, 2010

Last edited: Jan 18, 2010
2. Jan 18, 2010

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.