Problem 23 - Chemistry and Chemical Reactivity

Question

Forty miles above the earth's surface the temperature is \(250 \mathrm{K}\) and the pressure is only \(0.20 \mathrm{mm}\) Hg. What is the density of air (in grams per liter) at this altitude? (Assume the molar mass of air is \(28.96 \mathrm{g} / \mathrm{mol} .\) )

Step-by-Step Solution

Answer

The density of air at this altitude is approximately 0.000371 g/L.

Step 1: Convert Pressure to Atmospheres

First, we need to convert the pressure given in mm Hg to atmospheres because we will use the ideal gas law, which utilizes pressure in atmospheres. We know that 1 atm is equivalent to 760 mm Hg. Thus, to convert 0.20 mm Hg to atm, we use the formula:\[P_{atm} = \frac{0.20 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 2.63 \times 10^{-4} \, \text{atm}\]

Step 2: Use the Ideal Gas Law

The ideal gas law is expressed as \( PV = nRT \). We can manipulate this to find the density. First, recognize that \( n = \frac{m}{M} \), where \( m \) is the mass in grams and \( M \) is the molar mass, leading to \( PV = \frac{m}{M}RT \). Rearranging for density (\( \rho \)) gives:\[\rho = \frac{m}{V} = \frac{PM}{RT}\]

Step 3: Substitute Known Values

Now that we have the formula for density, substitute the known values into the equation:- \( P = 2.63 \times 10^{-4} \, \text{atm} \)- \( M = 28.96 \, \text{g/mol} \)- \( R = 0.0821 \, \text{L atm/mol K} \)- \( T = 250 \, \text{K} \)Substituting in these values:\[\rho = \frac{(2.63 \times 10^{-4}) \times 28.96}{0.0821 \times 250}\]

Step 4: Calculate the Density

Solving the expression from Step 3 gives:\[\rho = \frac{(2.63 \times 10^{-4}) \times 28.96}{0.0821 \times 250} = \frac{0.00761768}{20.525} \approx 0.000371 \text{ g/L}\]

Key Concepts

Pressure Conversion

In many scientific problems, pressures are given in various units like mm Hg or atmospheres. Knowing how to convert between these units is essential. The pressure in the original exercise was specified in millimeters of mercury (mm Hg), which is a common unit used in barometric readings. However, the Ideal Gas Law requires pressure to be in atmospheres (atm) to maintain consistency with the gas constant value.
To convert from mm Hg to atm, use the relation: 1 atm = 760 mm Hg. Thus, the calculation becomes a straightforward division. For example, converting 0.20 mm Hg to atm involves dividing by 760, resulting in approximately 2.63 × 10^-4 atm.
Mastering this conversion is fundamental when working with gases under varying conditions, ensuring accurate calculations with the Ideal Gas Law.

Gas Density

Gas density connects mass and volume, calculated via the Ideal Gas Law rearrangement. Typically, density is mass per unit volume (g/L). In the Ideal Gas Law, \[ PV = nRT \] where:

• P is pressure.
• V is volume.
• n is moles of gas.
• R is the gas constant.
• T is temperature in Kelvin.

Replacing moles with mass (m) divided by molar mass (M), the formula adapts to \[ \rho = \frac{PM}{RT} \].
This lets you link pressure, temperature, and gas identity (molar mass). In our altitude example, this calculation yielded a very low density, 0.000371 g/L, underscoring how altitude and lower pressure influence gas density.

Molar Mass

Molar mass is the mass of a mole of a substance, a crucial concept when dealing with gases. It allows us to calculate other properties like density through its inclusion in the Ideal Gas Law.
The exercise gives the molar mass of air as 28.96 g/mol. This average accounts for the common components of air, primarily nitrogen, oxygen, argon, and small amounts of other gases. Without this, finding the density wouldn't be possible because density calculation requires mass-to-volume ratios related back to the number of moles.
Understanding molar mass not only aids in direct exercises but also helps you when rearranging equations to solve for unknowns. Using molar mass accurately lets you transition from looking at matter in terms of quantities to real, measurable masses.