Question

If I have 4.1 moles of nitrous oxide (laughing gas) that is kept at 24.5°C in a container under 2.2 atm, then what is the volume of the container? [R = 0.0821 (L*atm)/(mol*K)]

Answer 1

We have a gas that we will assume behaves like an ideal gas, so we can apply the ideal gas equation.

$PV=nRT$

Where,

V is the volume of the gas

P is the pressure of the gas=2.2atm

T is the temperature of the gas = 24.5 + 273.15K=297.65K

R is a constan=0.0821atm.L/mol.K

n is the number of moles = 4.1moles

Now, we clear V and replace the known data:

$V=\frac{nRT}{P}$$\begin{gathered} V=\frac{4.1mol\times0.0821\frac{atm.L}{mol.K}\times297.65K}{2.2atm} \\ V=45.5L \end{gathered}$

The volume we find is the volume that the gas occupies. Now, gases due to their characteristics occupy the volume of the container where they are contained. Therefore, the volume of the container will also be 45.5L.

Answer: The volume of the container is 45.5L