Problem 23 - Chemistry and Chemical Reactivity

Question

Forty miles above Earth's surface, the temperature is \(250 \mathrm{K},\) and the pressure is only \(0.20 \mathrm{mm} \mathrm{Hg} .\) 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 is 0.000371 g/L at this altitude.

Step 1: Identify the Known Variables

We know the temperature (\(T\)) is \(250 \mathrm{K}\) and the pressure (\(P\)) is \(0.20 \mathrm{mm} \mathrm{Hg}\). The molar mass of air is \(28.96 \mathrm{g/mol}\). These will be important for calculating the density.

Step 2: Convert Pressure to Atmospheres

The ideal gas law uses pressure in atmospheres, so convert \(0.20 \mathrm{mm} \mathrm{Hg}\) to atmospheres. Use the conversion factor where \(1 \mathrm{atm} = 760 \mathrm{mm} \mathrm{Hg}\). Thus, \(P = \frac{0.20}{760} = 0.0002632 \mathrm{atm}\).

Step 3: Use Ideal Gas Law to Find Moles per Liter

The ideal gas law, \(PV = nRT\), can be rearranged to \(n/V = P/RT\) to find moles per liter. Use \(R = 0.0821 \mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K}\). Calculate:\[\frac{n}{V} = \frac{0.0002632 \mathrm{atm}}{0.0821 \mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K} \times 250 \mathrm{K}} = 1.281 \times 10^{-5} \mathrm{mol/L}\].

Step 4: Convert Moles per Liter to Grams per Liter

Multiply the moles per liter by the molar mass to convert to grams per liter. Calculate: \[\text{Density} = 1.281 \times 10^{-5} \mathrm{mol/L} \times 28.96 \mathrm{g/mol} = 0.000371 \mathrm{g/L}\].

Key Concepts

Density Calculation

Density is a measure of how much mass is contained in a given volume. For gases, density is often expressed in grams per liter (g/L). In the context of the Ideal Gas Law, density can be derived by rearranging the basic equation:

• The formula for density \( \rho \) can be linked to the Ideal Gas Law: \( PV = nRT \).
• By knowing that \( n = \frac{m}{M} \), where \( m \) is the mass and \( M \) is the molar mass, you can substitute to find density.
• This rearranges to \( \rho = \frac{PM}{RT} \). This equation shows that density is directly proportional to pressure and molar mass, and inversely proportional to temperature.

Use this equation for a quick calculation of the density of air at high altitude.

Molar Mass

Molar mass is the mass of a given substance (chemical element or chemical compound) divided by its amount of substance in moles. It's expressed in g/mol and represents the weight of one mole of that substance. For gases, molar mass is critical because:

• It allows you to convert between moles and grams, crucial for many calculations including those involving the Ideal Gas Law.
• In this exercise, the molar mass of air is considered as an average due to its composition of different gases; primarily nitrogen and oxygen.
• Using 28.96 g/mol as the molar mass of air is suitable for calculations related to the Earth's atmosphere.

Understanding molar mass helps in converting the molar volume of a gas to its mass.

Temperature Conversion

Temperature conversion isn't needed for the given exercise, since calculations were directly in Kelvin.

• The Kelvin scale is used in scientific calculations because it starts at absolute zero, the coldest possible temperature, which is vital for gas law calculations.
• This scale ensures there are no negative values, which simplifies using the Ideal Gas Law formula \( PV = nRT \).
• If temperature in Celsius was given, converting it to Kelvin would require adding 273.15 to the Celsius value.

Using the correct temperature scale is crucial for accurate results when applying scientific formulas.

Pressure Conversion

Pressure conversion is essential when using the Ideal Gas Law, as it requires pressure in atmospheres.

• Standard pressure units like mmHg and atm are used frequently. In this exercise, the pressure of 0.20 mmHg was converted to atm.
• Use the conversion factor of \( 1 \text{ atm} = 760 \text{ mmHg} \) to convert pressure.
• Converting to atm helps maintain consistency with the Ideal Gas Law constant \( R = 0.0821 \text{ L} \cdot \text{atm} / \text{mol} \cdot \text{K} \), which requires pressure in atm.

Converting to the correct pressure unit ensures calculations align with established gas constants for accurate results.