Questions: What is the density of Xe gas at 70°C and 2.50 atm?

Solution Steps

To find the density of a gas, we can use the Ideal Gas Law, which is given by:

\[ PV = nRT \]

where:

• \( P \) is the pressure,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant, and
• \( T \) is the temperature in Kelvin.

The temperature is given in degrees Celsius, so we need to convert it to Kelvin:

\[ T = 70 + 273.15 = 343.15 \, \text{K} \]

The density (\(\rho\)) of a gas is defined as mass per unit volume. We can express the number of moles \( n \) in terms of mass \( m \) and molar mass \( M \):

\[ n = \frac{m}{M} \]

Substituting this into the Ideal Gas Law:

\[ P = \frac{m}{M} \cdot \frac{RT}{V} \]

Rearranging for density \(\rho = \frac{m}{V}\):

\[ \rho = \frac{PM}{RT} \]

Given:

• \( P = 2.50 \, \text{atm} \)
• \( M = 131.293 \, \text{g/mol} \) (molar mass of Xe)
• \( R = 0.0821 \, \text{L atm/mol K} \)
• \( T = 343.15 \, \text{K} \)

Substitute these values into the density formula:

\[ \rho = \frac{2.50 \times 131.293}{0.0821 \times 343.15} \]

Calculate:

\[ \rho = \frac{328.2325}{28.169215} \approx 11.650 \, \text{g/L} \]

Final Answer