If a maximum of 5.60g of a gas at 2.50 atm of pressure dissolves in 3.5-L of water, what will the solubility be if the pressure changes to 5.00atm?

Question

aaronq · Answer

Assuming that the solubility is linear, since you're doubling the amount, double the mass will be soluble.

anonymous · Answer

shouldnt it still be 5.6g?

aaronq · Answer

um the way i'm thinking about this is like this|dw:1429844276294:dw|

since the gas pressure is higher, more should be dissolved into the liquid. An everyday example would be the gas dissolved in soda cans. Under pressure more gas is dissolved, when you open the can gas escapes from the liquid phase under the new reduced pressure.

anonymous · Answer

common sense gets the best of us.

anonymous · Answer

so, it'd be like:
$$
\frac{(5.6g/3.5L)}{2.5atm}=\frac{x}{5atm}\
x=5{\textrm atm}\times \frac{5.6 \textrm g}{2.5 {\textrm atm}\times3.5\textrm L}
\
x=\boxed{\quad} {\textrm g/L}$$
something in grams per litre ... ?

aaronq · Answer

yeah, that's the way I would imagine they want it solved. The relationship may not be linear in reality though, but the only way to test it would be to actually do the experiment.