What would be the mass of an O2 gas having a volume of 72L at a pressure of 22.7mmHg and a temperature of 87oC.

Question

What would be the mass of an O2 gas having a volume of 72L at a pressure of 22.7mmHg and a temperature of 87oC.

Answer 1

I arrived at a slightly different answer from @Joan L. It may have to do with the use of R and which value is used. If you leave the pressure in mm Hg and then use the corresponding value of R = 62.36 Ltorr/Kmol. I'm not quire sure which value of R was used in the previous answer. So here is my approach...

PV = nRT

n = moles O₂ = PV/RT

n = (22.7 mm Hg)(72 L) / (62.4 L-mmHg/Kmol)(360K)

n = 0.0728 moles O₂

mass O₂ = 0.0728 moles O₂ x 32 g/mol = 2.33 g O₂

Answer 2

Hello Jam,

You should have been exposed to the Ideal Gas Law which has many formats depending on the parameter you are trying to find.

Formats of the Ideal gas law are:

PV=nRT {n= # of moles of the gas}; PV=(g/M)RT {M = molar mass of the gas}; PM = dRT {d=density}

The format of the Ideal Gas Law to be used here is PV=(g/M)RT where M = molar mass of the gas, R= gas constant [760 (L.torr/mol.K) g = mass of the gas and T is the temperature in Kelvin ( which is ^o C +273).

The unit of pressure determines what value you use for the gas constant

R = 62.4 L.atm/mol.K =760 L.torr/mol.K [also remember that 1 torr =1 mm Hg]

Another important thing to watch for is that the units used will cancel eg. volume needed to be in liters since R has liters....

Using the format of the equation above

g = (PVM/RT) ; now insert the data provided giving (noting that the molar mass of oxygen is 32.0 g/mol)

g= [22.7 torr x 72 L x 32.0 g/mol ] / [760 L.torr/mol.K x 360 K] = 0.191 g O₂

I am aware that you might have already known some of the information shared here but shared anyway to complete your knowledge and review of the Ideal Gas Law.

All the best!