To find the molar mass of the gaseous compound, we'll use the Ideal Gas Law, rearranged to solve for the molar mass. The Ideal Gas Law is:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure of the gas,
- [tex]\( V \)[/tex] is the volume of the gas,
- [tex]\( n \)[/tex] is the number of moles of the gas,
- [tex]\( R \)[/tex] is the universal gas constant, and
- [tex]\( T \)[/tex] is the temperature of the gas.
We also know that the density [tex]\(\rho\)[/tex] is defined as:
[tex]\[ \rho = \frac{m}{V} \][/tex]
where:
- [tex]\(\rho\)[/tex] is the density,
- [tex]\( m \)[/tex] is the mass of the gas,
- [tex]\( V \)[/tex] is the volume.
For a given substance, the number of moles [tex]\( n \)[/tex] can be expressed as:
[tex]\[ n = \frac{m}{M} \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the gas,
- [tex]\( M \)[/tex] is the molar mass of the gas.
Substituting [tex]\( n = \frac{m}{M} \)[/tex] into the ideal gas equation, we have:
[tex]\[ P V = \frac{m}{M} RT \][/tex]
Rearranging to solve for [tex]\( M \)[/tex], we get:
[tex]\[ M = \frac{mRT}{PV} \][/tex]
Using the definition of density [tex]\(\rho = \frac{m}{V}\)[/tex], we can rewrite [tex]\( m \)[/tex] as [tex]\( \frac{\rho V}{1} \)[/tex]:
[tex]\[ M = \frac{\rho V RT}{PV} \][/tex]
Since [tex]\( V \)[/tex] cancels out, the equation simplifies to:
[tex]\[ M = \frac{\rho RT}{P} \][/tex]
Given the following values:
- Density [tex]\(\rho = 1.23 \text{ kg/m}^3\)[/tex]
- Temperature [tex]\( T = 330 \text{ K} \)[/tex]
- Pressure [tex]\( P = 20 \text{ kPa} = 20 \times 10^3 \text{ Pa} \)[/tex]
- Gas constant [tex]\( R = 8.314 \text{ J/(mol·K)} \)[/tex]
We plug in these values into our formula:
[tex]\[ M = \frac{1.23 \times 8.314 \times 330}{20 \times 10^3} \][/tex]
(Note: The pressure needs to be converted from kPa to Pa, which means multiplying by 1000.)
After calculating this, we find:
[tex]\[ M \approx 0.16873263 \; \text{kg/mol} \][/tex]
So, the molar mass of the compound is approximately [tex]\( 0.16873263 \; \text{kg/mol} \)[/tex].
