Final answer:
The volume of CO2 that would diffuse through the same hole in the same time can be calculated using Graham's law of effusion, which relates the rates of diffusion of two gases to their molar masses.
Step-by-step explanation:
The question relates to Graham's law of effusion, which describes the rate at which gases diffuse or effuse through a barrier. Using this law, we can correlate the rate of diffusion of one gas to another if we know their molecular masses. In the case given, 32 ml of hydrogen diffuses through a fine hole in 1 minute, and we are asked to calculate what volume of CO₂ will diffuse in 1 minute.
Let's denote the rate of diffusion of hydrogen as R₁ and CO₂ as R₂. According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (M).
r1/r2 = √(M2/M1)
Graham's law formula can be rewritten to solve for the volume of CO₂ (V₂) given the volume of hydrogen diffused (V₁):
V2 = V1 * √(M1/M2)
Where V₁ is the volume of hydrogen, M1 is the molar mass of hydrogen (approximately 2 g/mol), M2 is the molar mass of CO₂ (approximately 44 g/mol), and V₂ is the volume of CO₂ we want to find. Substituting the values, we have:
V2 = 32 ml * √(2 g/mol / 44 g/mol)
After calculating the square root and multiplying, we would get the volume of CO₂ that diffuses in one minute. The key assumption here is that conditions such as temperature and pressure are constant for both gases during their diffusion.