The actual diameter of an atom is about a tenth of a nanometer (10^-10 m). In or
ID: 980776 • Letter: T
Question
The actual diameter of an atom is about a tenth of a nanometer (10^-10 m). In order to develop some intuition for the molecular scale of a gas, assume that you are considering a liter of air(mostly N2 and O2) at room temperature and a pressure of one atmosphere -- 10^5 Pa. A. Calculate the number of molecules in the sample of gas. B. Estimate the average spacing between the molecules. C. Estimate the average speed of a molecule using the relationship between the KE of a molecule and the temperature. D. Suppose that the gas were rescaled upwards so that each atom was the size of a tennis ball. What would be the average spacing between molecules? The actual diameter of an atom is about a tenth of a nanometer (10^-10 m). In order to develop some intuition for the molecular scale of a gas, assume that you are considering a liter of air(mostly N2 and O2) at room temperature and a pressure of one atmosphere -- 10^5 Pa. A. Calculate the number of molecules in the sample of gas. B. Estimate the average spacing between the molecules. C. Estimate the average speed of a molecule using the relationship between the KE of a molecule and the temperature. D. Suppose that the gas were rescaled upwards so that each atom was the size of a tennis ball. What would be the average spacing between molecules? The actual diameter of an atom is about a tenth of a nanometer (10^-10 m). In order to develop some intuition for the molecular scale of a gas, assume that you are considering a liter of air(mostly N2 and O2) at room temperature and a pressure of one atmosphere -- 10^5 Pa. A. Calculate the number of molecules in the sample of gas. B. Estimate the average spacing between the molecules. C. Estimate the average speed of a molecule using the relationship between the KE of a molecule and the temperature. D. Suppose that the gas were rescaled upwards so that each atom was the size of a tennis ball. What would be the average spacing between molecules?Explanation / Answer
PV = nRT
n = PV/RT
n = 1atm * 1 L / 0.0821 * 298 K
n = 0.04087 moles
number of molecules = 0.04087 * 6.023 * 10^23
= 2.462 * 10^22
Average kinetic energy K.E = 3/2 kT
= 3/2 *(( 2.462 * 10^22)/(0.04087 * 8.314))*298
= 3.239 * 10^25
volume of the gas molecules = (2.462 * 10^22)*(( 10^-10 m))^3
= 2.462 * 10^-8 m3
= 2.462 * 10^-5 L
empty space between gas molecules = 1 - (2.462*10^-5) = 0.99997 L
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