Question

Calculate the pressure in a 212 liter tank containing 23.3 kg of argon gas at 25°c

Answer 1

First of all, let's find the number of moles of the gas.

The molar mass of argon is $M_m=40 g/mol=0.40 kg/mol$. Since we have $m=23.3 kg$ of gas, the number of moles is
$n= \frac{m}{M_m}= \frac{23.3 kg}{0.40 kg/mol}=58.3 mol$

Now we can use the ideal gas law to calculate the pressure of the gas:
$pV=nRT$
where
p is the pressure
$V=212 L=0.212 m^3$ is the volume
$n=58.3 mol$ is the number of moles
$R=8.31 J/mol K$ is the gas constant
$T=25^{\circ}+273=298 K$ is the absolute temperature

Rearranging the equation, we find
$p= \frac{nRT}{V}= \frac{(58.3 mol)(8.31 J/mol K)(298 K)}{0.212 m^3}=6.81 \cdot 10^5 Pa$