Question

If 0.100 mol of argon gas occupies 2.15 L at \(725 \mathrm{~mm} \mathrm{Hg}\), what is the temperature in degrees Celsius?

Short answer

The temperature is approximately -23.35 °C.

Step-by-step solution

Identify the Known Variables

We have the following known values: the number of moles of argon gas, \(n = 0.100\) mol; the volume \(V = 2.15\) L; and the pressure \(P = 725 \mathrm{~mm} \mathrm{Hg}\). We'll convert the pressure to atm for calculation purposes: \(1 \mathrm{~atm} = 760 \mathrm{~mm} \mathrm{Hg}\), so \(P = \frac{725}{760}\, \mathrm{atm} \approx 0.9539 \mathrm{~atm}\).

Recall the Ideal Gas Law

The ideal gas law formula is \(PV = nRT\), where \(R\) is the ideal gas constant \(R = 0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\). Our goal is to find \(T\), the temperature in Kelvin first.

Rearrange the Ideal Gas Law for T

To solve for \(T\), we rearrange the formula as follows: \(T = \frac{PV}{nR}\).

Plug in the Known Values

Substitute \(P = 0.9539 \mathrm{~atm}\), \(V = 2.15 \mathrm{~L}\), \(n = 0.100 \mathrm{~mol}\), and \(R = 0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) into the formula: \[ T = \frac{0.9539 \times 2.15}{0.100 \times 0.0821} \approx \frac{2.050885}{0.00821} \approx 249.80 \mathrm{~K} \].

Convert Temperature from Kelvin to Celsius

To convert the temperature from Kelvin to degrees Celsius, use the relation \(T_{\mathrm{C}} = T_{\mathrm{K}} - 273.15\). Substitute \(T_{\mathrm{K}} = 249.80 \mathrm{~K}\): \[ T_{\mathrm{C}} = 249.80 - 273.15 \approx -23.35 \mathrm{~°C} \].

Key concepts

Temperature conversion

Temperature conversion is essential in scientific calculations, especially when dealing with different systems, like Kelvin and Celsius scales. The Kelvin scale is often used in gas law problems because it starts from absolute zero, making calculations straightforward.
To convert from Kelvin to Celsius, you subtract 273.15 from the Kelvin temperature. This conversion is crucial because while the Kelvin scale is needed for calculations, Celsius is often more intuitive for everyday use.
For instance, a temperature of 249.80 K converts to Celsius by the formula: \[ T_{\mathrm{C}} = T_{\mathrm{K}} - 273.15 \]So, 249.80 K becomes \( T_{\mathrm{C}} = 249.80 - 273.15 = -23.35 \mathrm{~°C} \).

• The Kelvin scale is used for its convenience in scientific formulas, like the Ideal Gas Law.
• Celsius, on the other hand, is more user-friendly for everyday temperature references.

Pressure conversion

Converting pressure units is often necessary when dealing with gas laws, as different units may be used in different contexts. In this example, pressure was initially given in mm Hg, a common unit in older scientific literature and medicine.
However, the Ideal Gas Law calculations require pressure in atmospheres (atm) for consistency with the gas constant R, which is 0.0821 L atm mol\(^{-1}\) K\(^{-1}\). To convert mm Hg to atm:

• Recognize that 1 atm equals 760 mm Hg.
• Divide the pressure in mm Hg by 760 to get pressure in atm.\[ P = \frac{725}{760} \approx 0.9539 \ \mathrm{atm} \]

This conversion makes the calculation compatible with the Ideal Gas Law.

Gas laws

The Ideal Gas Law is a foundational equation in chemistry that relates pressure, volume, temperature, and the number of moles of a gas. This law is expressed as \[ PV = nRT \]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.

This equation assumes ideal conditions where gas particles do not interact, and the volume of gas particles is negligible. It's versatile for solving various problems involving gases, like finding unknown variables when others are known.
In our example, the known values of pressure, volume, and moles allow for solving the unknown temperature, demonstrating the law’s utility.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. While it wasn't directly part of the exercise provided, stoichiometry often overlaps with gas laws when gases are involved in reactions.

• Molecular stoichiometry can help determine the amount of gas produced or consumed in a reaction using the Ideal Gas Law.
• Understanding mole-to-mole relationships is key, as is converting gases from volume to moles using the gas law formula, \( n = \frac{PV}{RT} \), which allows calculations to involve gases similarly to solids and liquids.

This helps with calculating yields, limiting reactants, and other factors in chemical equations and processes, therefore integrating gas concepts with chemical reactions.