Question

What is the volume occupied by \[11g\] of carbon dioxide at \[27^\circ C\] and \[780mm\] of $Hg$ pressure?

Answer 1

Hint: Charles is one of the important laws in the gaseous state. Charles' law was proposed by J.A.C. Charles. This law is used to study the relationship volume of a gas and its temperature. In Charles' law the mass of the system and pressure of the system is constant. The Combination of Charles law, Boyle's law and Avogadro’s hypothesis is known as the ideal gas equation.
Formula used:
Moles are defined as the given mass of the molecule is divided by the molecular mass of the molecule.
\[{\text{moles}}{\text{ = }}\dfrac{{{\text{mass}}{\text{of molecule}}}}{{{\text{molecular weight of the molecule}}}}\]
The molecular weight of the molecule is dependent on the atomic weight of the atom present in the molecule. The molecular weight of the molecule is equal to the sum of the molecular weight of the atom and the number of the respective atoms in the molecule.
\[\text{Molecular Weight} = \text{Number of atoms} \times \text{Atomic Weight Of The Atom}\]
The ideal gas equation depends on the pressure, temperature, number of moles, volume of the gas molecules in ideal condition.
The ideal gas equation is,
\[PV = nRT\]
Here, the pressure of the gas is P
The volume of the gas is V
The temperature of the gas in kelvin is T
Gas constant is R
The number of moles of the Gas molecules is n
Formula for convert degree Celsius to kelvin in temperature
\[kelvin \]= \[ degree + 273\]

Complete answer:
Given,
Temperature at degree = \[27^\circ C\]
Temperature at kelvin = \[27 + 273 = 300k\]
Gas constant, R=\[R = 0.08206Latm/kmol\]
The pressure of the gas is \[780mm\].
 The carbon dioxide has one carbon atom and two oxygen atoms in the molecular formula.
The atomic weight of carbon is \[12g\].
The atomic weight of oxygen is \[16g\].
The molecular mass of carbon dioxide is,
$ = 12 + \left( {16 \times 2} \right)$
$ = 12 + 32 = 44g$
The number of moles of carbon dioxide present in \[11g\],
The given mass of carbon dioxide is \[11g\]
\[moles = \dfrac{{\text{mass of molecule}}}{{\text{molecular weight of the molecule}}}\]
\[moles = \dfrac{{11}}{{44}} = 0.25moles\]
The number of moles is \[0.25\] moles.
The ideal gas equation is
\[PV = nRT\]
We change the formula for calculate the volume of the gas,
\[V = \dfrac{{nRT}}{P}\]
We substitute the known values in formula
\[V = \dfrac{{0.25 \times 0.08206 \times 300 \times 760}}{{780}} = 6litre\]
The volume occupied by \[11g\] of carbon dioxide at \[27^\circ C\] and \[780mm\] of $Hg$ pressure is \[6litre\].
According to the above discussion and calculation, we conclude the volume occupied by \[11g\]of carbon dioxide at \[27^\circ C\] and \[780mm\] of $Hg$ pressure is \[6litre\].

Note:
We have to know that the air is nothing but a mixture of the gases. The mixture of the gas in the air are major in the form of nitrogen, oxygen, carbon dioxide and other gases. In the atmosphere approximately \[78\% \] of nitrogen, \[21\% \] of oxygen and remaining \[1\% \] of other gases in the world. The symbol of nitrogen is \[N\] and oxygen is \[O\]. The natural form of nitrogen and oxygen gases are diatomic. The diatomic nature of nitrogen and oxygen gases are \[{N_2}\] and \[{O_2}\].