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= (m)/(M_m)= (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= (nRT)/(V)= ((58.3 mol)(8.31 J/mol K)(298 K))/(0.212 m^3)=6.81 \cdot 10^5 Pa`