Question

Assume that the volume of a fixed amount of gas in a rigid container does not change. Calculate the pressure the gas would exert if the temperature were changed as shown in the following table. $$ \begin{array}{|c|c|c|c|} \hline \begin{array}{c} \text { Initial } \\ \text { Pressure } \end{array} & \begin{array}{c} \text { Initial } \\ \text { Temperature } \end{array} & \begin{array}{c} \text { Final } \\ \text { Temperature } \end{array} & \begin{array}{c} \text { Final } \\ \text { Pressure } \end{array} \\ \hline 302 \text { torr } & 0.0^{\circ} \mathrm{C} & 105.0^{\circ} \mathrm{C} & ? \\ \hline 735 \text { torr } & 25.0^{\circ} \mathrm{C} & 0.0^{\circ} \mathrm{C} & ? \\ \hline 3.25 \mathrm{~atm} & 273 \mathrm{~K} & 373 \mathrm{~K} & ? \\ \hline \end{array} $$

Short answer

The final pressures for the three gases are approximately 418.8 torr, 673.6 torr, and 4.86 atm respectively.

Step-by-step solution

Convert initial temperatures to Kelvin

To convert the initial temperatures from degree Celsius to Kelvin, to each of the temperatures add 273.15. Thus, for the three gases, the initial temperatures in Kelvin become: \n\n1) 273 K\n2) 298 K (25 + 273 K)\n3) 273 K

Convert final temperatures to Kelvin

Perform the same conversion for the final temperatures as follows: \n\n1) 378 K \n2) 273 K \n3) 373 K

Use Gay-Lussac's law to calculate the final pressure

Using the formula \(P_1/T_1 = P_2/T_2\), rearrange the terms to solve for \(P_2\) to get \(P_2 = \frac{P_1 \cdot T_2}{T_1}\). Substituting each value of initial pressure, final temperature, and initial temperature into the formula, calculate the final pressure: \n\n1) \(P_2 = \frac{302 \cdot 378}{273} =\) 418.8 torr approx \n2) \(P_2 = \frac{735 \cdot 273}{298} =\) 673.6 torr approx \n3) \(P_2 = \frac{3.25 \cdot 373}{273} =\) 4.86 atm approx.

Key concepts

Pressure-Temperature Relationship

The pressure-temperature relationship explains how gas pressure changes with temperature. When a gas's temperature increases, its particles move more vigorously, colliding with each other and the container's walls more forcefully.
As temperature goes up, so does pressure, assuming the volume remains constant. If the temperature decreases, particles move less energetically, and the pressure drops. This relationship is critical in predicting and calculating changes in gas pressure in response to temperature fluctuations.
The mathematical relationship is depicted in Gay-Lussac's Law, which we'll cover in a bit more detail further down. Understanding this relationship is essential for applications both in everyday life and scientific endeavors. Consider that car tires can change pressure with temperature as an example.
Everyday applications include:

• Car tire pressure changes.
• Cooking at different altitudes.
• Managing pressurized gas containers.

Kelvin Temperature Scale

The Kelvin scale is integral when dealing with gas laws, because it starts at absolute zero, the point where molecular motion stops.
In the gas laws, we express temperatures in Kelvin to ensure calculations reflect the direct proportionality between temperature and pressure or volume. The Kelvin scale is incremental like the Celsius scale but shifted by 273.15 degrees.
So, converting from Celsius to Kelvin is simple: add 273.15.
Let's look at examples for clarity:

• 0°C becomes 273.15 K.
• 25°C translates to 298.15 K.

Using Kelvin ensures the math aligns correctly, reinforcing the theoretical foundations of gas laws like Gay-Lussac's law.
Whenever you're working within these calculations, always remember: zero Kelvin (-273.15°C) is the lowest possible temperature, marking the absence of all thermal energy.

Gay-Lussac's Law

Gay-Lussac's Law highlights the direct proportionality between pressure and temperature for a gas with a fixed volume.
The law is mathematically represented as \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), 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.
This means that if you increase the temperature, the pressure will also increase, assuming the volume does not change.
In our exercise, we use Gay-Lussac's Law to predict the final pressures based on initial conditions of temperature and pressure.

• The pressure in a vessel increases as the temperature rises.
• If the temperature drops, the pressure decreases accordingly.

When applying this law, always make sure temperatures are in Kelvin to maintain the consistency of direct proportionality. This law is crucial in understanding not only everyday occurrences but also in more complex scientific scenarios.