Question

A mercury lamp contains \(0.0055 \mathrm{~g}\) of Hg vapor in a volume of \(15.0 \mathrm{~mL}\). If the operating temperature is 2,800 K, what is the pressure of the mercury vapor?

Short answer

The pressure of the mercury vapor is approximately 0.419 atm.

Step-by-step solution

Identify Known Quantities

We are given the mass of mercury vapor as \(0.0055\, \text{g}\), the volume of the container as \(15.0\, \text{mL}\), and the temperature as \(2800\, \text{K}\). We aim to find the pressure of the mercury vapor.

Convert Mass to Moles

First, we need to convert the mass of mercury from grams to moles using the molar mass of mercury, which is approximately \(200.59\, \text{g/mol}\). The number of moles \(n\) is calculated as:\[ n = \frac{0.0055\, \text{g}}{200.59\, \text{g/mol}}\approx 2.74 \times 10^{-5}\, \text{mol}.\]

Convert Volume to Liters

The volume given is in milliliters, so we need to convert it to liters for use in the ideal gas law. \(15.0\, \text{mL}\) is \(0.015\, \text{L}\).

Apply Ideal Gas Law

Use the ideal gas law, \(PV = nRT\), where \(R\) is the ideal gas constant \(0.0821\, \text{L}\, \text{atm}\, \text{mol}^{-1}\, \text{K}^{-1}\). Rearrange to solve for pressure \(P\):\[P = \frac{nRT}{V} = \frac{2.74 \times 10^{-5}\, \text{mol} \times 0.0821\, \text{L atm mol}^{-1} \text{K}^{-1} \times 2800\, \text{K}}{0.015\, \text{L}}.\]

Calculate the Pressure

Perform the calculation:\[P = \frac{2.74 \times 10^{-5} \times 0.0821 \times 2800}{0.015} \approx 0.419\, \text{atm}.\] Thus, the pressure of the mercury vapor is approximately \(0.419\, \text{atm}\).

Key concepts

Molar Mass of Mercury

To solve problems involving substances like mercury vapor, knowing its molar mass is essential. The molar mass of a substance is the mass of one mole of its molecules or atoms. For mercury, the molar mass is approximately 200.59 grams per mole (g/mol). To convert a given mass of mercury to moles, you can use the formula:

• Number of moles \( n = \frac{\text{mass of mercury}}{\text{molar mass of mercury}} \).

For example, if you have 0.0055 grams of mercury, you would calculate the moles by dividing the mass by the molar mass. Here it would be:\[ n = \frac{0.0055\, \text{g}}{200.59\, \text{g/mol}} \approx 2.74 \times 10^{-5}\, \text{mol}. \]This conversion is crucial in the application of the ideal gas law, which involves the number of moles of the gas in calculations. Understanding molar mass allows you to link the macroscopic measurements we make in the lab to molecular scale information.

Gas Pressure Calculation

Calculating gas pressure essentially involves using the ideal gas law equation, which is a fundamental equation in chemistry: \( PV = nRT \). In this formula:

• \(P\) is the pressure of the gas in atmospheres (atm).
• \(V\) is the volume of the gas in liters (L).
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant, approximately 0.0821 L atm/mol K.
• \(T\) is the temperature in Kelvin (K).

To find the pressure \(P\), you would rearrange the equation to solve for pressure:\[ P = \frac{nRT}{V}\]Take your calculated number of moles from the molar mass conversion, plug in the known volume, temperature, and the ideal gas constant into the equation. For instance, given 2.74 x 10^-5 moles, 0.015 L volume, and 2800 K temperature:\[ P = \frac{2.74 \times 10^{-5} \times 0.0821 \times 2800}{0.015} \approx 0.419\, \text{atm}.\]This calculation gives you the pressure exerted by the mercury vapor in the lamp, providing insights into its behavior under those conditions.

Temperature Conversion in Chemistry

Understanding how to convert temperatures is an important skill in chemistry, especially when dealing with gas laws. The Kelvin scale is commonly used in these calculations because it starts at absolute zero, where particles theoretically stop all motion. This makes it a convenient scale for theoretical calculations.If you are dealing with a temperature given in Celsius and need to convert it to Kelvin, the conversion is straightforward. You merely add 273.15 to the Celsius temperature:

• \( T_{K} = T_{C} + 273.15 \)

For example, if you had a temperature of 26.85°C, the conversion to Kelvin would be:\[ T_{K} = 26.85 + 273.15 = 300\, \text{K}.\]Always ensure that the temperature in any ideal gas law calculation is in Kelvin, as the gas constant \(R\) is formulated for this scale. Maintaining consistency across units is critical to obtaining correct results in chemical calculations.