QUESTION:
A vessel contains a mixture of nitrogen (m_1 = 7.0\rm \ g) and carbon dioxide (m_2 = 11\rm \ g) at a temperature T = 290\rm \ K and pressure P = 1.0\rm \ atm. Find the density of this mixture, assuming the gases to be ideal.
To solve this problem, you must have an idea of the ideal gas law.
SOLUTION:
Let:
- n is the number of moles of the mixture,
- \rho is the density of the mixture,
- M_1 is the gram molecular mass of nitrogen (\rm N_2),
- M_2 is the gram molecular mass of carbon dioxide (\rm CO_2), and
- P , V and T are the pressure, volume and temperature of the mixture respectively.
Given:
- m_1 = 7.0\rm \ g
- m_2 = 11\rm \ g
- T = 290\rm \ K
- P = 1.0\rm \ atm
We know the values
- R = 0.0821\rm \ L.atm.mol^{-1}.K^{-1}
- M_1 = 28.0\rm \ g
- M_2 = 44.0\rm \ g
The total number of moles of the gasses present in the mixture is
\begin{aligned}n &= \left (\dfrac{m_1}{M_1}+\dfrac{m_2}{M_2} \right )\\&=\left (\dfrac{7.0\rm \ g}{28.0\rm \ g}+\dfrac{11\rm \ g}{44.0\rm \ g} \right )\\&=0.5\\\end{aligned}
From the ideal gas equation
\begin{aligned}PV &= nRT\\\Rightarrow V &=\dfrac{nRT}{P}\\& = \dfrac{(0.5)(0.0821\rm \ L•atm•mol^{-1}K^{-1})(290\rm \ K)}{( 1.0\rm \ atm)}\\&= 11.9045\rm \ L\\&= 11.9045 \times 10^{-3}\rm \ m^3 \\\end{aligned}
Now, we have the total mass of the mixture, m = m_1+m_2 = 18\rm \ g = 18\times 10^{-3}\rm \ kg, and volume of the mixture, V= 11.9045 \times 10^{-3}\rm \ m^3.
Therefore, the density of the mixture can be calculated as
\rho = \dfrac{m}{V} = \dfrac{18\times 10^{-3}\rm \ kg }{ 11.9045 \times 10^{-3}\rm \ m^3} = 1.5\rm \ kg/m^3
