Question

A sample of pure gas has a density of ${\text{1}}{\text{.60 gm/L}}$ at a temperature of $26.5^{o}\,C$and pressure of $680.2{\text{ mm Hg}}$. Which of the following gas could be the sample?
A. $C{H_4}$
B. ${C_2}{H_6}$
C. $C{O_2}$
D. $Xe$

Answer 1

To solve this question we will find the molecular mass of the sample by using Gas law. All gas particles obey Gas law. From Gas law we know that,
$P = \dfrac{d}{M}RT$where,
$P = {\text{ Pressure}} \\\ d = {\text{ density}} \\\ M = {\text{ Molar mass}} \\\ R = {\text{ Universal Gas Constant}} \\\ T = {\text{Temperature}} \\\$

Complete step by step answer:
Given data:
Density of gas = ${\text{1}}{\text{.60 gm/L}}$
Temperature of gas = $26.5^{o}\,C$ = $299.5\,K$ and,
Pressure of gas = $680.2{\text{ mm Hg = }}\dfrac{{680.2}}{{760}}{\text{ mm}}$
$R = 0.0821{\text{ d}}{{\text{m}}^3}{\text{ atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}{\text{ }}$
We know that from gas law,
$P = \dfrac{d}{M}RT$,
Substituting the given data in above equation we get,
$P = \dfrac{d}{M}RT \\\ P = {\text{ Pressure}} \\\ d = {\text{ density}} \\\ M = {\text{ Molar mass}} \\\ R = {\text{ Universal Gas Constant}} \\\ T = {\text{Temperature}} \\\ \Rightarrow M = \dfrac{{dRT}}{P} \\\ \Rightarrow M = \dfrac{{1.6 \times 0.0821 \times 299.5}}{{\dfrac{{680.2}}{{760}}}} \\\ \Rightarrow M = 44 \\\$
In the given option molecular weight of $C{O_2}$ is 44.
Therefore, $C{O_2}$​ is the correct answer.
Hence, option C is correct.

Additional Information:
Gas laws:
-Boyle’s law: this law relates volume and pressure of a given gas at constant temperature. In this law he stated that “at a constant temperature the volume of gas is inversely proportional to pressure”.
$V \propto \dfrac{1}{P}$ (At a constant temperature)
-Charles’ law: this relates the volume and temperature of a given mass of gas at a constant pressure. In this law he stated that “at a constant pressure the volume is directly proportional to temperature”.
$V \propto T$
-Pressure-Temperature law: this law relates pressure and temperature of a given mass of gas at a constant volume. It states that “at a constant volume, pressure of the gas increases or decreases by 1/273 of its pressure at $0^\circ C$ per degree change of temperature.
${P_1} = {P_0} + \dfrac{{{P_0} \times t}}{{273}}$

Note: $R$ can be expressed in different units but for pressure- volume calculations. R must be taken in the same units of pressure and volume. And other unit conversions should also be taken care before substituting in the equation.