Problem 23 - Chemistry and Chemical Reactivity

Question

Forty miles above Earth's surface, the temperature is \(250 \mathrm{K},\) and the pressure is only \(0.20 \mathrm{mm}\) 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 0.0037 g/L.

Step 1: Convert Pressure Units

First, we need to express the given pressure from mm Hg to atm for calculation convenience. Use the conversion \(1\; ext{atm} = 760\; ext{mm Hg}\). Thus, \(0.2 \; ext{mm Hg} = \frac{0.2}{760}\; ext{atm}\).

Step 2: Use Ideal Gas Law

Apply the ideal gas law formula \(PV = nRT\). We rewrite it as \(P = \frac{n}{V}RT\) to compute density. This becomes \(P = \frac{m}{M}RT\), where \(m\) is the mass of air, \(M\) the molar mass, and \(\frac{m}{V}\) the density \(d\). Therefore, \(d = \frac{PM}{RT}\).

Step 3: Insert Known Values

Insert the known values into the density equation: \(d = \frac{{0.000263157\; ext{atm}} \times 28.96\; ext{g/mol}}{0.0821\; ext{L atm/mol K} \times 250\; ext{K}}\).

Step 4: Calculate Density

Perform the calculation: \(d = \frac{0.000263157 \times 28.96}{0.0821 \times 250}\). Simplifying gives \(d \approx 0.0037 \; ext{g/L}\).

Key Concepts

Density Calculation

Calculating the density of a gas involves understanding the relationship between mass, volume, and molar mass. The concept of density can be defined as the mass of a substance per unit volume. For gases, we use the ideal gas law to derive a formula for density.The ideal gas law is given by the equation:\[ PV = nRT \]where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.We can manipulate this equation to express density. - We know that \( n = \frac{m}{M} \), indicating the number of moles is the mass divided by the molar mass. - Substituting this into the ideal gas law gives: \[ PV = \frac{m}{M}RT \]- Rearranging gives an expression for density \( d = \frac{m}{V} \): \[ d = \frac{PM}{RT} \]This formula helps calculate the density of air or any other gas when pressure and temperature are known. Simply insert the values for pressure, molar mass, the ideal gas constant, and temperature to find the density.

Molar Mass

Molar mass is a critical concept for understanding many chemical calculations, including those involving gases. It is the mass of a given substance (molecular or atomic) divided by the amount of substance in moles.For air, the average molar mass is approximately 28.96 g/mol, which results from the mix of gases present in the air, mainly nitrogen and oxygen.When calculating properties like gas density, molar mass is crucial because it allows us to relate moles to mass:- The formula \( n = \frac{m}{M} \) shows that the number of moles \( n \) is the quotient of the mass \( m \) and molar mass \( M \).Thus, in the context of the density calculation or ideal gas law:- Molar mass acts as a bridge converting between the mass of the substance and the number of moles.

Understanding molar mass not only aids in density calculations but also provides insight into how different gases will behave under similar conditions, making it a cornerstone for further studies in chemistry.

Pressure Conversion

Converting pressure units is often necessary when dealing with gases and gas laws, particularly in chemistry and physics. In the exercise, the pressure is given in mm Hg, a common unit of pressure, but calculations involving gases, especially using the ideal gas law, are often carried out in atmospheres (atm).To convert pressure from mm Hg to atm, the known relation is:- \( 1\; \text{atm} = 760\; \text{mm Hg} \)Thus, to convert from mm Hg to atm, you can use the formula:\[ P_{\text{atm}} = \frac{P_{\text{mm Hg}}}{760} \]This conversion is necessary because the ideal gas constant \( R \) is often given in \( \text{L atm/mol K} \), which requires pressure in atm.The ideal gas constant standardizes calculations, and using different pressure units without conversion would lead to inaccuracies. Therefore, understanding pressure conversion is essential for accurate application of the ideal gas law and related calculations.