How many kilograms of #O_2# gas are needed to give an oxygen pressure of 3.48 atm at 22°C?

Question

Ellie Andrews · Accepted Answer

Just to retire this question, let us specify a volume of #1*m^3#; and I get a mass of dioxygen gas of under #5*kg#.

We use the Ideal Gas Equation, with #R=0.0821*L*atm*K^-1*mol^-1#. And thus #n=(PV)/(RT)=(3.48*cancel(atm)xx1*cancel(m^3)xx10^3*cancelL*cancel(m^-3))/(0.0821*cancel(L*atm)*cancel(K^-1)*mol^-1xx295*cancelK)# #=143.7*mol# (note that the answer has units #1/(mol^-1)=1/(1/(mol))=mol# #"as required"# And we change this molar amount to a mass by: #143.7*cancel(mol)xx32.00*cancel(g*mol^-1)xx10^-3*kg*cancel(g^-1)# #=4.60*kg.# I would like you to check my math. There are no refunds. Note that #1*m^3# is a very large volume, and is equal to #1000*L#. Chemists tend to deal with #"litres"# and #mL# and #cm^3# (#1*mL-=1*cm^3-=10^-3*L-=10^-6*m^3#). Of course all the measurements should be equivalent dimensionally, but it is so easy to divide instead of multiply and vice versa.

Cameron Andrews · Answer

undefined To calculate the mass of O₂ gas, we can use the ideal gas law:

$$PV = nRT$$

Where:
- $P$ is pressure (in atm),
- $V$ is volume (in liters),
- $n$ is the number of moles,
- $R$ is the ideal gas constant (0.0821 L.atm/(mol.K)),
- $T$ is temperature (in Kelvin).

Rearranging the formula to solve for moles ($n$):

$$n = \frac{PV}{RT}$$

To find the mass ($m$), we use the molar mass of O₂ (32 g/mol):

$$m = n \times \text{molar mass of } O_2$$

Given values:
- $P = 3.48 \, \text{atm}$,
- $V$ is not provided,
- $R = 0.0821 \, \text{L.atm/(mol.K)}$,
- $T = 22°C = 295 \, \text{K}$.

If you have the volume ($V$), you can substitute these values into the formula to find the number of moles ($n$), and then calculate the mass ($m$).