Question

What is the density of helium at ${{500}^{\circ }}C$ and 100mm pressure ?

Answer 1

To solve this question we should be aware of ideal gas, ideal gas equation and gas density. Ideal gases are gases that satisfy Boyl's law, Charle's law and Avagadro's law. Helium gas is an ideal gas.

Complete Solution :
Ideal gas are the gas that obey charle's law, Boyle's law and avagadro 's law. If gas deviate from these law will be real gas.
We know that helium is a ideal gas
The ideal gas equation is:
PV= nRT……………equation 1
where, P = Pressure
V = Volume
n = Number of moles
R = Gas constant
T = Temperature
Rearranging equation 1
$\dfrac{n}{V}=\dfrac{P}{RT}$……equation 2

We know that , n = $\dfrac{mass\text{ }of\text{ }gas\text{ }\left( g \right)~~~}{molar\text{ }mass\text{ }\left( g/mol \right)}$ ……………equation 3
substituting equation 3 in equation 2
$\dfrac{mass\text{ }of\text{ }gas\text{ }~~~}{molar\text{ }mass\times V}=\dfrac{P}{RT}$…………………equation 4

we know that, Density = $\dfrac{mass\text{ }of\text{ }gas~~}{V}$……………..equation 5
Substituting equation 5 in 4
d = $\dfrac{PM}{RT}$……………………………….equation 6
Where, d = gas density
M = molar mass of the gas
Equation 6 is the required formula to calculate density of helium,
Let's solve the given problem,
Given: P = 100mm
= $\dfrac{100}{760}$ as 1atm =760mm
= 0.1315atm
M = 4
R = 0.0821
T = ${{500}^{\circ }}C$
= 773.15K (as ${{0}^{\circ }}C$ = 273.15K)

Now, substituting the above values in equation 6
d =$\dfrac{PM}{RT}$
d = $\dfrac{0.1315\times 4}{0.0821\times 773.14}$
= $\dfrac{0.526}{63.4747}$
= 0.008286

The density always should be g/cc
so,= 0.008286$\times {{10}^{3}}$
= 8.286g/cc
Thus, the density of helium is 8.286g/cc.

Note: Always write the unit of all the quantity mentioned because not following this may lead to the wrong answer. We should be aware of deriving gas density from ideal gas equations as sometimes they might twist the question. It is very important to know the ideal gas.