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Chapter 10: Problem 80

Suppose you have two 1 -L flasks, one containing \(\mathrm{N}_{2}\) at STP, the other containing \(\mathrm{CH}_{4}\) at STP. How do these systems compare with respect to (a) number of molecules, (b) density, (c) average kinetic energy of the molecules, (d) rate of effusion through a pinhole leak?

Short Answer

Expert verified
a) Both N2 and CH4 have the same number of molecules in a 1-L flask at STP. b) N2 is denser than CH4 (\(1.25\,\text{g/L}\) vs \(0.713\,\text{g/L}\)). c) The average kinetic energy of the molecules is the same for both N2 and CH4 at STP. d) The rate of effusion of CH4 is greater than that of N2.

Step by step solution

01

a) Number of molecules

First, we need to find the number of moles of each gas in a 1-L flask at STP. Since 1 mole of an ideal gas occupies 22.4 L at STP, the number of moles in a 1-L flask is given by: \(n = \frac{1 \text{ L}}{22.4 \text{ L/mol}} = 0.04464 \text{ mol}\) Using Avogadro's number, we convert the moles to the number of molecules: \(\text{molecules} = n \times Avogadro's\, number\) \(\text{molecules} = 0.04464 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol}\) The number of molecules in each flask is the same, because the number of moles is the same for both N2 and CH4 at STP.
02

b) Density

To find the density, we need to determine the mass of each gas with the given molecular weights. Density = mass / volume For N2: \(\rho_\text{N2} = \frac{0.04464\,\text{mol} \times 28\,\text{g/mol}}{1\,\text{L}} = 1.25\,\text{g/L}\) For CH4: \(\rho_\text{CH4} = \frac{0.04464\,\text{mol} \times 16\,\text{g/mol}}{1\,\text{L}} = 0.713\,\text{g/L}\) So, N2 is denser than CH4.
03

c) Average kinetic energy

The formula for average kinetic energy (Ek) of a molecule in a gas: \(Ek = \frac{3}{2}kT\) Where k is the Boltzmann constant, k = \(1.38 \times 10^{-23} J/K\), and T is the temperature in Kelvin. For both N2 and CH4, the Ek is the same at STP: \(Ek = \frac{3}{2}(1.38 \times 10^{-23}\,\text{J/K}) \times 273.15\,\text{K}\) The average kinetic energy of the molecules is the same for both N2 and CH4 at STP.
04

d) Rate of effusion

To find the rate of effusion, we use Graham's law: \(Rate_1 / Rate_2 = \sqrt{M_2 / M_1}\) Comparing N2 (Rate_1) and CH4 (Rate_2): \(\frac{\text{Rate}_\text{N2}}{\text{Rate}_\text{CH4}} = \sqrt{\frac{16\,\text{g/mol}}{28\,\text{g/mol}}}\) Thus, the rate of effusion of CH4 is greater than that of N2. In summary: a) The number of molecules is the same for both N2 and CH4 at STP. b) N2 is denser than CH4. c) The average kinetic energy of the molecules is the same for both N2 and CH4 at STP. d) The rate of effusion of CH4 is greater than that of N2.

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