Question

A typical gas cylinder used for such depths contain 69.33g of O2 and 457.24g of N2 and has a volume of 20.00L. What is the partial pressure of each gas at 15 degrees, and what is the total pressure in the cylinder at this temperature?

MM(O2)= 32.00 u
MM(N2)=28.0u
R= 0.0821L

Partial pressure of O2=
Partial pressure of N2=
Total pressure in the tank=

Answer 1

In calculating the number of moles of nitrogen and oxygen gas, we use the ideal gas law;

$n=$ $PV\over RT$

$nO_2=$ $(1atm)(0.06933L)\over (0.0821{atm.L\over mol.K})(273+15k)$ $=2.9\times 10^{-3} mol O_2$

$nN_2=$ $(1atm)(0.45724L)\over (0.0821L{atm.L\over mol.L})(273+15k)$

$=1.9\times 10^{-2}mol N_2$

In calculating the partial pressure of the gases, we use Daltons equation

$P=$ $nRT\over V$

$P_{O_2}=$ $(2.9\times 10^{-3})(0.0821{atm.L\over mol.L})(288k)$

$=0.0685atmO_2$

$P_{N_2}=(1.9\times 10^{-2})(0.0821{atm.L\over mol.L})(288k)$

$=0.449atm N_2$

$P_{Total}=P_{N_2}+P_{O_2}$

$=0.449atm+0.0685atm$

$P_{Total}=0.5175atm$