Please help me with this chemistry question.

Question

A volume of 200 mL of a gas are collected by water displacement at a temperature of 30 degrees Celsius and a pressure of 105 kPa. Given that the vapour pressure of water at 30 degrees celsius is 4.5 kPa, calculate the volume the gas would occupy at SATP.

Answer 1

There's a lot going on in this question so let's break it down. First, we know this is an ideal gas law type of question, but what are we trying to solve for? The question asks what volume the gas would occupy at STP. Something that's useful to have memorized is the volume of an ideal gas at STP is 22.4 L per mol of gas. Therefore, if we can figure out how many moles of gas were collected via water displacement, then we can figure out its volume at STP.

Whenever you collect gas via water displacement, the gas will be a mixture of whatever you collected plus water vapor. This means the total pressure of the mixture will be sum of the partial pressure of water plus the partial pressure of the gas (Dalton's Law of Partial Pressure). The partial pressure of the water is equal to its vapor pressure. Therefore, if the total pressure of the mixture is 105 kPa, then the partial pressure of the gas is 100.5 kPa (after subtracting off the partial pressure of water vapor).

Now we have a volume (200 mL), a temperature (30 C), and a pressure (100.5 kPa). With these, we can use PV = nRT (ideal gas law) to solve for n, the number of moles. However, we cannot just blindly plug in these numbers into the equation. We must convert our units.

200 mL = 0.200 L

30 C = 303.15 K

100.5 kPa = 0.9919 atm

R = 0.0821 L*atm/mol*K

Rearranging our equation to solve for n, we get:

n = PV/RT

n = (0.9919)(0.200)/(0.0821)(303.15)

n = 0.00797 moles of gas

Now we can figure out the volume at STP. If you don't remember that each mole has a volume of 22.4 L, then we can simply use the ideal gas law again to calculate the volume.

Option A: 0.00797 mol * 22.4 L/mol = 0.179 L

Option B: V = nRT/P

V = (.00797)(.0821)(273)/(1) = 0.179 L

Finally, let's check to see if our answer makes sense. To get to STP, we had to increase the pressure (from 0.99 atm initially to 1 atm) and decrease the temperature (from 303.15 K to 273.15 K). Increase pressure and decreasing temperature both will decrease the volume (according to Boyle's and Charles' Law respectively). Thus, our volume at STP should be smaller than our initial volume. Since we started at 0.200 L and ended at 0.179 L, our answer is consistent with what we would expect conceptually.