Answer :
Answer:
To find the rate of diffusion of carbon dioxide (CO2), we can use Graham’s Law of Effusion. This law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass. The formula is given by:
Rate of diffusion of Gas 2Rate of diffusion of Gas 1=Molar mass of Gas 1Molar mass of Gas 2
Given that the rate of diffusion of nitrogen (N2) is 0.04 cm/s, and the molar masses of nitrogen (N2) and carbon dioxide (CO2) are approximately 28 g/mol and 44 g/mol respectively, we can calculate the rate of diffusion of CO2 as follows:
Rate of diffusion of CO20.04 cm/s=2844
Solving for the rate of diffusion of CO2 gives us:
Rate of diffusion of CO2=28440.04 cm/s
Rate of diffusion of CO2=1.57140.04 cm/s
Rate of diffusion of CO2=1.25360.04 cm/s
Rate of diffusion of CO2=0.0319 cm/s
Therefore, the rate of diffusion of carbon dioxide (CO2) is approximately 0.0319 cm/s.
Required answer:
- 0.032 cm/s
Explanation:
According to "Graham's law of diffusion", the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Here we are given, A gas mixture contains Nitrogen (N2) and Carbon dioxide (CO2) in a 3:1 molar ratio. If the rate of diffusion of nitrogen is 0.04 cm/s.
Then according to Graham's law of diffusion
[tex]\sf \dfrac{Rate \ of \ diffusion \ of \ N_2}{Rate \ of \ diffusion \ of \ CO_{2}} = \sqrt{\dfrac{Molar \ Mass \ of \ N_2}{Molar \ Mass \ of \ CO_{2}}}[/tex]
- Molar mass of N2 = 28 g/mol
- Molar mass of CO2 = 44 g/mol
Substitute the values,
[tex] \sf\dfrac{0.4 \ cm/s}{R_{CO_{2} }} = {\sqrt{\dfrac{28}{44}}} \\ \\ \sf \dfrac{0.4 cm/s}{R_{CO_{2}}} = \sqrt{0.636} \\ \\ \sf{R_{CO_{2}} = \sqrt{0.636} \times 0.4 } \\ \\ \sf{R_{CO_{2}} = 0.032 \ cm/s}[/tex]
[tex]\therefore [/tex] The rate of diffusion of carbon dioxide is 0.032 cm/s
