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

AI Thread Summary
The discussion focuses on calculating the work done by an ideal gas during the reaction of tin dissolving in acid. The user applies the formula for change in volume, ΔV, based on the change in moles of hydrogen gas produced. They derive the work done using the equation w = -PΔV, leading to w = -(2 mol H2)RT. However, they express uncertainty about the accuracy of their calculations since they cannot verify the answer against their textbook. The conversation highlights the importance of understanding gas laws in chemical reactions.
AdkinsJr
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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.
 
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