Naked Science Forum
On the Lighter Side => Science Experiments => Topic started by: scientizscht on 28/04/2020 21:03:26

Hello!
How can I convert kPa (pressure) into molarity? I do not have any other information e.g. volume, I only have the molecular weight of the gas.
Thanks!

Assume that you have a large room full of gas and it has the same mass as a small donkey.
Then do the calculation.
You may be surprised by the result.

Assume that you have a large room full of gas and it has the same mass as a small donkey.
Then do the calculation.
You may be surprised by the result.
What calculation?

22.4 liters of gas at STP equals 1 mole.

So what is the molarity of a 50kPa oxygen partial pressure?

Assume that you have a large room full of gas and it has the same mass as a small donkey.
Then do the calculation.
You may be surprised by the result.
What calculation?
The one where you calculate the molarity based on the information you have, together with the volume of a large room, and the mass of a small donkey.

So what is the molarity of a 50kPa oxygen partial pressure?
That question doesn't make sense. It is like asking, "What is the volume water if it's density is 1.01 gm/cubic cm?"

So what is the molarity of a 50kPa oxygen partial pressure?
That question doesn't make sense. It is like asking, "What is the volume water if it's density is 1.01 gm/cubic cm?"
It does make sense.
Much more like asking what's the volume of a mole of water if its density is 1 g/ cc
(and with a bigger temperature dependence, but I'm going to assume it's 20C)

What's the calculation then? Is there any assumption regarding partial pressures like they refer to 1L?

The only assumption you need is the temperature.

but I'm going to assume it's 20C)

OK I have some oxygen.
It has the same mass as a donkey, and occupies a large room.
I don't know what that is so I will call it M (in grams).
And I don't know what volume it has, so I will call that V (in litres).
And I am assuming it's at 20 C or 293K
And I have the pressure it's 50 KPa That's close enough to half atmospheric pressure, so I'm going to call it 0.5
I know it has a molecular weight of 32, because it's oxygen.
So I have M/32 moles of O2.
PV=nRT
And n = M/32
So P V=MRT /32
Divide through by M
PV/M =RT/32
Divide through by P
V/M = RT/(32P)
Take the reciprocal
M/V= 32P/(RT)
OK that tells me M/V which is the density of the gas.
That's how many grams I have in a litre
And I was asked for the molarity which is the number of moles in a litre.
So I need to convert mass to moles no problem 1 gram is 1/32 moles
So the number of moles per litre is
(M/32) /V = 32 P/RT
Interestingly the 32 cancels
I can look up R
8.31446261815324×10−2 L⋅bar⋅K−1⋅mol−1
From wiki
https://en.wikipedia.org/wiki/Gas_constant
(it's important to match the units)
Call it 0.0831 L⋅bar⋅K−1⋅mol−1
Molarity (in moles per litre) =P/ RT
P is 0.5 bar (near enough)
P/RT =0.5/(0.0831 *293)
So, the molarity is 0.0205 moles per litre
No donkeys required.
Of course, any chemist would be less of an ass.
A mole of any gas at room temperature and 1 atmosphere occupies about 24 litres
https://en.wikipedia.org/wiki/Molar_volume#Ideal_gases
So a litre is 1/24 moles ( for 1 atmosphere)
so any gas has a concentration (at 1 atm) of 1/24 moles/litre
But this gas is at 0.5 atm so there's only half as much gas.
any gas at 0.5 atm has a concentration of 1/48 moles per litre
1/48 =0.0208 molar.
But where's the fun in that?
All you need to know is the temperature and pressure, and you can calculate the molarity of a gas without knowing the molecular mass.
Which is why Bob said this
22.4 liters of gas at STP equals 1 mole.

So what is the molarity of a 50kPa oxygen partial pressure?
That question doesn't make sense. It is like asking, "What is the volume water if it's density is 1.01 gm/cubic cm?"
It does make sense.
Much more like asking what's the volume of a mole of water if its density is 1 g/ cc
(and with a bigger temperature dependence, but I'm going to assume it's 20C)
In fairness, I was not wrong, you had to assume a value for one of the variables to answer the question.
I did like your analysis.

So what do you think, is the test complete.