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Chapter 5: Problem 29

In Example 5.11 of the text, the molar volume of \(\mathrm{N}_{2}(g)\) at \(\mathrm{STP}\) is given as 22.42 \(\mathrm{Lmol} \mathrm{N}_{2} .\) How is this number calculated? How does the molar volume of He(g) at \(\mathrm{STP}\) compare to the molar volume of \(\mathrm{N}_{2}(g)\) at \(\mathrm{STP}\) (assuming ideal gas behavior)? Is the molar volume of \(\mathrm{N}_{2}(g)\) at 1.000 atm and \(25.0^{\circ} \mathrm{C}\) equal to, less than, or greater than 22.42 \(\mathrm{L} / \mathrm{mol} ?\) Explain. Is the molar volume of \(\mathrm{N}_{2}(g)\) collected over water at a total pressure of 1.000 \(\mathrm{atm}\) and \(0.0^{\circ} \mathrm{C}\) equal to, less than, or greater than 22.42 \(\mathrm{L} / \mathrm{mol}\) ? Explain.

Short Answer

Expert verified
The molar volume of N2(g) at STP is 22.42 L/mol, which is the same as the molar volume of He(g) at STP due to their ideal gas behavior. At 1.0 atm and 25.0°C, the molar volume of N2(g) is greater than 22.42 L/mol, specifically 24.47 L/mol. When collected over water at 1.0 atm and 0.0°C, the molar volume of N2(g) is also greater than 22.42 L/mol, calculated as 22.59 L/mol.

Step by step solution

01

Calculate the molar volume of N2(g) at STP

To calculate the molar volume of a gas at STP (Standard Temperature and Pressure), we use the Ideal Gas Law: \(PV=nRT\) where P is pressure in atm, V is volume in liters, n is the number of moles of the gas, R is the ideal gas constant, and T is temperature in Kelvin. At STP, the pressure is 1 atm and the temperature is 273.15 K. In this case, we have 1 mole of N2(g), so we can rearrange the Ideal Gas Law to solve for V: \(V = \frac{nRT}{P}\) Substitute the values for STP and the gas constant (R = \(0.0821 \frac{L \cdot atm}{K \cdot mol}\)): \(V = \frac{(1 \ mol)(0.0821 \frac{L \cdot atm}{K \cdot mol})(273.15 \ K)}{(1 \ atm)} = 22.42 \ L/mol\) The molar volume of N2(g) at STP is 22.42 L/mol.
02

Compare the molar volume of He(g) at STP to that of N2(g)

Ideal gases at the same temperature and pressure have the same molar volume, regardless of the identity of the gas. Since He(g) and N2(g) are both ideal gases at STP, their molar volumes are the same: Molar volume of He(g) at STP = Molar volume of N2(g) at STP = 22.42 L/mol
03

Molar volume of N2(g) at 1.0 atm and 25.0°C

To determine if the molar volume of N2(g) at 1.0 atm and 25.0°C is equal to, less than, or greater than 22.42 L/mol, we can calculate the molar volume at these conditions using the Ideal Gas Law. First, convert the temperature to Kelvin: 25.0°C + 273.15 = 298.15 K Now, calculate the molar volume: \(V = \frac{(1 \ mol)(0.0821 \frac{L \cdot atm}{K \cdot mol})(298.15 \ K)}{(1 \ atm)} = 24.47 \ L/mol\) The molar volume of N2(g) at 1.0 atm and 25.0°C is greater than 22.42 L/mol.
04

Molar volume of N2(g) collected over water at 1.0 atm and 0.0°C

When a gas is collected over water, its total pressure includes the vapor pressure of water. To find the partial pressure of the gas, we can subtract the vapor pressure of water at 0.0°C (which is 0.00604 atm) from the total pressure: Partial pressure of N2(g) = 1.0 atm - 0.00604 atm = 0.99396 atm Now, calculate the molar volume of N2(g) at 0.0°C (which is 273.15 K) and the partial pressure: \(V = \frac{(1 \ mol)(0.0821 \frac{L \cdot atm}{K \cdot mol})(273.15 \ K)}{(0.99396 \ atm)} = 22.59 \ L/mol\) The molar volume of N2(g) collected over water at 1.0 atm and 0.0°C is greater than 22.42 L/mol.

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Most popular questions from this chapter

Cyclopropane, a gas that when mixed with oxygen is used as a general anesthetic, is composed of 85.7\(\% \mathrm{C}\) and 14.3\(\% \mathrm{H}\) by mass. If the density of cyclopropane is 1.88 \(\mathrm{g} / \mathrm{L}\) at STP, what is the molecular formula of cyclopropane?

For scuba dives below 150 \(\mathrm{ft}\) , helium is often used to replace nitrogen in the scuba tank. If 15.2 \(\mathrm{g}\) of \(\mathrm{He}(g)\) and 30.6 \(\mathrm{g}\) of \(\mathrm{O}_{2}(g)\) are added to a previously evacuated 5.00 \(\mathrm{L}\) tank at \(22^{\circ} \mathrm{C},\) calculate the partial pressure of each gas present as well as the total pressure in the tank.

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You have an equimolar mixture of the gases \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2},\) along with some \(\mathrm{He}\), in a container fitted with a piston. The density of this mixture at STP is 1.924 \(\mathrm{g} / \mathrm{L}\) . Assume ideal behavior and constant temperature and pressure. a. What is the mole fraction of He in the original mixture? b. The \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) react to completion to form \(\mathrm{SO}_{3} .\) What is the density of the gas mixture after the reaction is complete?

Consider separate \(1.0-\mathrm{L}\) samples of \(\mathrm{He}(g)\) and \(\mathrm{UF}_{6}(g),\) both at 1.00 atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?
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