Question

The density of a gas at $760mm$ pressure and $O^{o}C$ temperature is $32$. What will be its density at $570mm$ procedure, if the temperature remains constant?

Answer 1

Solution:-

From ideal gas equation-

$PV=nRT$

whereas,

$P=$ pressure of the gas

$V=$ volume it occupies

$n=$ number of moles of gas present in the sample $=\frac{\text{mass of gas}\left(m\right)}{\text{molar mass of gas}\left(M\right)}$

$R=$ universal gas constant $=0.0821atmL{mol}^{-1}{K}^{-1}$

$T=$ absolute temperature of the gas

$\therefore PV=\frac{m}{M}RT$

$\Rightarrow PM=\frac{m}{V}RT$

As we know that,

$\frac{m}{V}=\rho \left(\text{density}\right)$

$\therefore PM=\rho RT$

For constant temperatur,

Density of a gas is directly proportional to the pressure, i.e.,

$\rho \propto P$

$\therefore \frac{\rho }{{\rho }^{\prime }}=\frac{P}{P^{\prime }}$

$\Rightarrow {\rho }^{\prime }=\frac{\rho \times P^{\prime }}{P}$

Given that:-

$P=760mm$

$\rho =32g{L}^{-1}$

$P^{\prime }=570mm$

${\rho }^{\prime }=?$

$\therefore {\rho }^{\prime }=\frac{32\times 570}{760}=24g{L}^{-1}$

Hence, density at $570mm$ pressure will be $24g{L}^{-1}$ .