# Question ca88b

Oct 11, 2016

Q1

Given Molar mass of the gas $146 \text{ g/mol}$

By Ideal gas law we know

$P V = \frac{w}{M} R T$

where

$P \to \text{Pressure of the gas} = 1 a t m$

$V \to \text{Volume of the gas}$

$R \to \text{Universal gas constant} = 0.082 L a t m {K}^{-} 1 m o {l}^{-} 1$

$T \to \text{Temperature of the gas in K} = \left(150 + 273\right) K = 423 K$

$w \to \text{Mass of the gas}$

$M \to \text{Molar mass of the gas"=146" g/mol}$

$P V = \frac{w}{M} R T$

$\implies P = \frac{\frac{w}{V}}{M} R T = \frac{D}{M} \times R T$

$\text{ where " D="Density of the gas at given temperature and pressure}$
So
$D = \frac{P M}{R T}$

Inserting values we have

$\implies D = \frac{P M}{R T} = \frac{1 \times 146}{0.082 \times 423} = 4.21 \frac{g}{L}$

Q2

By ideal gas law we have

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2. \ldots . . \left(2\right)$

Given
${P}_{1} \to \text{Initial pressure} = 73 c m = \frac{73}{760} a t m$

${T}_{1} \to \text{Initial Temperature} = {68}^{\circ} F = x K$

$\frac{x - 273}{373 - 273} = \frac{68 - 32}{212 - 32}$

$\implies x = \frac{36}{180} \times 100 + 273 = 293 K$

${V}_{1} \to \text{Initial Volume} = 875 m L = 0.875 L$

${P}_{2} \to \text{Final pressure} = 12 a t m$

${T}_{2} \to \text{Finial Temperature} = 350 K$

V_2->"Final Volume"=?#

Inserting values in equation (2)

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

$\implies \frac{{P}_{2} {V}_{2}}{T} _ 2 = \frac{{P}_{1} {V}_{1}}{T} _ 1$

$\implies {V}_{2} = \frac{{P}_{1} {V}_{1} {T}_{2}}{{P}_{2} {T}_{1}}$

$\implies {V}_{2} = \frac{\frac{73}{760} \times 0.875 \times 350}{12 \times 293} L \approx 0.0084 L = 8.4 m L$