Question

A 1.00-L sample of argon gas at 1.00 atm is heated from \(20^{\circ} \mathrm{C}\) to \(110^{\circ} \mathrm{C}\). If the volume remains constant, what is the final pressure?

Short answer

The final pressure of the gas is approximately 1.31 atm.

Step-by-step solution

Understand the Problem

We have a sample of argon gas in a rigid container, meaning the volume does not change. The initial temperature is \(20^{\circ} \mathrm{C}\) and the final temperature is \(110^{\circ} \mathrm{C}\). We need to find the final pressure after heating, using the initial pressure of 1.00 atm.

Convert Temperature to Kelvin

Since we are dealing with gas laws, temperatures must be in Kelvin. Convert the initial temperature \(20^{\circ} \mathrm{C}\) to Kelvin by adding 273.15: \(293.15 \mathrm{K}\). Convert the final temperature \(110^{\circ} \mathrm{C}\) to Kelvin: \(383.15 \mathrm{K}\).

Apply Gay-Lussac's Law

Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature when volume is held constant. The formula is \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\), where \(P_1\) and \(P_2\) are initial and final pressures, and \(T_1\) and \(T_2\) are initial and final temperatures in Kelvin.

Calculate Final Pressure

We know \(P_1 = 1.00\, \text{atm}\), \(T_1 = 293.15\, \text{K}\), \(T_2 = 383.15\, \text{K}\). Rearrange the formula to solve for \(P_2\): \(P_2 = P_1 \times \frac{T_2}{T_1}\). Substitute the known values: \(P_2 = 1.00 \times \frac{383.15}{293.15}\).

Calculate and Report the Result

After calculation, we find \(P_2 \approx 1.31\, \text{atm}\). Thus, the final pressure of the gas at \(110^{\circ} \mathrm{C}\) is approximately 1.31 atm.

Key concepts

Gas Laws

Gas laws are fundamental principles in chemistry that describe the behavior of gases in relation to temperature, volume, and pressure. Understanding these laws enables us to predict how changes in one of these variables can affect the others under certain conditions.

There are several gas laws, but one of the key ones in this context is Gay-Lussac's Law. Gay-Lussac's Law states that if the volume of a gas is held constant, the pressure of the gas is directly proportional to its absolute temperature. This means that as the temperature increases, the pressure also increases, provided the volume does not change.

When using Gay-Lussac's Law in calculations, remember that temperatures should always be in Kelvin. The formula used is \[\frac{P_1}{T_1} = \frac{P_2}{T_2}\] where \(P_1\) and \(P_2\) are the initial and final pressures, and \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin. Utilizing this formula helps solve problems involving changes in pressure due to temperature variations.

Temperature Conversion to Kelvin

When working with gas laws, especially those like Gay-Lussac's that involve temperature, it's crucial to convert the temperatures from Celsius to Kelvin. This is because the Kelvin scale is an absolute scale with its zero point set at absolute zero, the point at which particles have minimum thermal motion.

To convert a temperature from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, an initial temperature of \(20^{\circ} \text{C}\) becomes \(293.15 \text{K}\) when converted, while \(110^{\circ} \text{C}\) converts to \(383.15 \text{K}\).

Using Kelvin is essential as it allows us to apply gas laws correctly, ensuring that our calculations and interpretations of a gas's behavior under varying temperatures and pressures are accurate.

Pressure Calculations

Calculating pressure changes in a gas involves using the proper formula to understand how variables like temperature affect pressure. In scenarios where volume is held constant, such as in rigid containers, Gay-Lussac's Law becomes particularly useful.

To calculate the final pressure when a gas is heated, we arrange Gay-Lussac's equation to solve for \(P_2\):\[P_2 = P_1 \times \frac{T_2}{T_1}\]Where \(P_1\) is the initial pressure, \(T_1\) is the initial temperature in Kelvin, \(T_2\) is the final temperature in Kelvin, and \(P_2\) is the final pressure.

By substituting known values into the formula, we can determine \(P_2\). For instance, starting with \(P_1 = 1.00\, \text{atm}\), \(T_1 = 293.15\, \text{K}\), and \(T_2 = 383.15\, \text{K}\), our calculation yields a \(P_2\) of approximately \(1.31\, \text{atm}\). Understanding this concept allows us to predict how pressure will change under different temperature conditions while the gas is confined in a fixed volume.