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Question: What is the volume occupied by \[11g\] of carbon dioxide at \[27^\circ C\] and \[780mm\] of \(Hg\) p...

What is the volume occupied by 11g11g of carbon dioxide at 27C27^\circ C and 780mm780mm of HgHg pressure?

Explanation

Solution

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.
moles = massof moleculemolecular weight 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.
Molecular Weight=Number of atoms×Atomic Weight Of The Atom\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=nRTPV = 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
kelvinkelvin = degree+273 degree + 273

Complete answer:
Given,
Temperature at degree = 27C27^\circ C
Temperature at kelvin = 27+273=300k27 + 273 = 300k
Gas constant, R=R=0.08206Latm/kmolR = 0.08206Latm/kmol
The pressure of the gas is 780mm780mm.
The carbon dioxide has one carbon atom and two oxygen atoms in the molecular formula.
The atomic weight of carbon is 12g12g.
The atomic weight of oxygen is 16g16g.
The molecular mass of carbon dioxide is,
=12+(16×2)= 12 + \left( {16 \times 2} \right)
=12+32=44g= 12 + 32 = 44g
The number of moles of carbon dioxide present in 11g11g,
The given mass of carbon dioxide is 11g11g
moles=mass of moleculemolecular weight of the moleculemoles = \dfrac{{\text{mass of molecule}}}{{\text{molecular weight of the molecule}}}
moles=1144=0.25molesmoles = \dfrac{{11}}{{44}} = 0.25moles
The number of moles is 0.250.25 moles.
The ideal gas equation is
PV=nRTPV = nRT
We change the formula for calculate the volume of the gas,
V=nRTPV = \dfrac{{nRT}}{P}
We substitute the known values in formula
V=0.25×0.08206×300×760780=6litreV = \dfrac{{0.25 \times 0.08206 \times 300 \times 760}}{{780}} = 6litre
The volume occupied by 11g11g of carbon dioxide at 27C27^\circ C and 780mm780mm of HgHg pressure is 6litre6litre.
According to the above discussion and calculation, we conclude the volume occupied by 11g11gof carbon dioxide at 27C27^\circ C and 780mm780mm of HgHg pressure is 6litre6litre.

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%78\% of nitrogen, 21%21\% of oxygen and remaining 1%1\% of other gases in the world. The symbol of nitrogen is NN and oxygen is OO. The natural form of nitrogen and oxygen gases are diatomic. The diatomic nature of nitrogen and oxygen gases are N2{N_2} and O2{O_2}.