# The ideal gas equation is $pV=nRT$. Prove that $p=\frac{\rho RT}{M}$

The ideal gas equation is $$pV=nRT$$

Show that the pressure exerted by a fixed mass of gas is given by the equation $$p=\frac{\rho RT}{M}$$

Where $$\rho$$ is the density of the gas and $$M$$ is the mass of one mole of gas.

Proof: $$\rho = \frac{m}{V}$$ where $$m,V$$ is mass and volume of the gas respectively.

$$p=\frac{nRT}{V}=\frac{n\rho RT}{m}$$

But we know that, $$m=nM \implies \frac{n}{m}=\frac{1}{M}$$

$$\therefore p=\frac{\rho RT}{M}$$

(b) The Earth’s atmosphere may be treated as an ideal gas whose density, pressure and temperature all decrease with height.

In 1924, Howard Somervell and Edward Norton set a new altitude record when attempting to climb Mount Everest. They managed to climb to a vertical height of $$8570 m$$ above sea level by breathing in natural air. At this height, the air pressure was $$0.35$$ times the pressure at sea level and the temperature was $$−33 °C$$. At sea level, air has a temperature $$20 °C$$ and density $$1.3~kg~m^{-3}$$

Calculate the density of the air at a height of $$8570 m$$ at the time the record was set.

As I was just asked to prove the above equation, I am going to assume that it will be useful for this part of the problem.

So, we have (at sea level) $$p=\frac{1.3 \times R \times 293}{M_1}$$

and at our height of $$8570m$$, $$p=\frac{\rho \times R \times 240}{M_2}$$

Now onto my question. Would $$M's$$ in my two equations be equal? I.e. would the mass of one mole of air at sea level and at our height of $$8750m$$ be equal? If so, could I have an explanation why please?