HYdrogen is collected by water displacement in a gas collecting tube. THe gas colume is 25.0 mL. Levels of?

Question

...water, inside and outisde the gas collecting tube, are the same. Temperature is 17 degrees celsius. Barometric pressure is 720.0 mmHg. Convert the volume to that of dry gas at Standard temp and pressure (STP).

the answer is 21.8 mL but can someone explain please?
THanks

Answer 1

First subtract the vapour pressure of water at 25 C from 720.0. That gives you P in torr. Then use

P1V1/T1 = P2V2/T2

Use whatever units for P you like,as long as they are the same on both sides. You can leave V in mL. T must of course be in K.

If you neglect to subtract the vapour pressure of water, you will get too high a volume.

Answer 2

T1 = 17 + 273 = 290 K
p1 = 720 / 760 = 0.947 atm
V1 = 0.0250 L
At STP
T2 = 273 K
p2 = 1.0 atm
Use the combined gas law
p1V1/T1 = p2V2/T2
0.947 x 0.0250 / 290 = 1.0 x V2 / 273
V2 = 0.0223 L = 22.3 mL

Answer 3

STP gas conditions are
T = 298 K
P = 760 mmHg
According to the IDEAL gas LAW
(PV / T)o = nR = K = (PV / T)
i.e.
(720 x 25 / 290) = (760 x V / 298)
V = 24.3 mL
But
This CALCUL is not taking into account the
1) gas SOLUBILITY (in the water) by increasing T
2) ideality DEVIATION
So, knowing the first, it is possible to obtain the right equation that represents the REALity.