Answer to Question #251439 in Molecular Physics | Thermodynamics for Bebe

Question

A large mining company was provided with a 3 cu. meters of compressed air tank. Air pressure in the tank drops from 700 kPa to 180 kPa while the temperature remains unchanged at 28 ⁰C. What percentage has the mass of air in the tank been reduced.

Answer 1

Ideal gas law

$pV=nRT \\ pV= \frac{m}{M}RT \\ m = \frac{pVM}{RT}$

Initial mass:

$m_i = \frac{p_iVM}{RT}$

Final mass:

$m_f = \frac{p_fVM}{RT} \\ Δm=m_i -m_f = (p_i- p_f) \frac{VM}{RT}$

Percantage mass reduced:

$= \frac{Δm}{m_i} \times 100 = \frac{(p_i-p_f) \frac{VM}{RT}}{p_i \frac{VM}{RT}} \times 100 \\ = \frac{p_i -p_f}{p_i} \times 100 \\ = \frac{700-180}{700} \times 100 \\ = 74.2 \; \%$