Question

Consider a container separated into two compartments by a barrier. One compartment has a gas at a pressure of 1 atm and the other compartment is empty. The barrier has a small hole in it, which allows the gas to diffuse from one compartment to another. If the rate of diffusion of the gas is 0.2 mL/min and the molar mass of the gas is 28g/mol, what is the mass of gas transferred through the hole in 5 hours?

Answer 1

First, let's determine the total volume of gas that diffuses through the hole in 5 hours.Rate of diffusion = 0.2 mL/minTime = 5 hours = 5 * 60 min = 300 minTotal volume diffused = Rate of diffusion * TimeTotal volume diffused = 0.2 mL/min * 300 min = 60 mLNow, we need to convert the volume of gas to moles. To do this, we'll use the Ideal Gas Law:PV = nRTWhere P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. We are not given the temperature, but since the gas is at 1 atm pressure, we can assume that it is at standard temperature and pressure  STP . At STP, 1 mole of any gas occupies 22.4 L  or 22,400 mL .P = 1 atm  pressure V = 60 mL  volume R = 0.0821 L atm/mol K  ideal gas constant T = 273.15 K  standard temperature Rearranging the Ideal Gas Law to solve for n:n = PV / RTFirst, we need to convert the volume from mL to L:V = 60 mL *  1 L / 1000 mL  = 0.06 LNow, we can plug in the values:n =  1 atm * 0.06 L  /  0.0821 L atm/mol K * 273.15 K n  0.0027 molesNow that we have the number of moles, we can find the mass of the gas transferred through the hole. We are given the molar mass of the gas as 28 g/mol.Mass = moles * molar massMass = 0.0027 moles * 28 g/mol  0.0756 gSo, the mass of gas transferred through the hole in 5 hours is approximately 0.0756 grams.