Question

A container is filled with an ideal gas to a pressure of 11.0 atm at \(0^{\circ} \mathrm{C} .\) a. What will be the pressure in the container if it is heated to \(45^{\circ} \mathrm{C} ?\) b. At what temperature would the pressure be 6.50 atm? c. At what temperature would the pressure be 25.0 atm?

Short answer

The solutions are: a. The pressure at \(45^{\circ}C\) will be about 12.76 atm; b. The temperature for a pressure of 6.50 atm would be approximately \(-144.61^{\circ}C\); c. The temperature for a pressure of 25.0 atm would be approximately \(348.49^{\circ}C\).

Step-by-step solution

Convert Temperatures to Kelvin

First, we need to convert Celsius temperatures to Kelvin for all given temperatures. Initial temperature, \(T_1\): \(T_1 = 0^{\circ} C + 273.15 = 273.15 K\) Temperature for part (a), \(T_2\): \(T_2 = 45^{\circ} C + 273.15 = 318.15 K\)

Use Gay-Lussac's Law to Solve Each Part

a. Calculate the final pressure, \(P_2\), when temperature is increased to \(45^{\circ} C\): We can rearrange the formula to solve for the final pressure, this gives us: \(P_2 = P_1 \times \frac{T_2}{T_1}\) Substitute known values: \(P_2 = 11.0 \, atm \times \frac{318.15 K}{273.15 K} \) Now, calculate the pressure: \(P_2 ≈ 12.76\, atm\) b. What temperature would the pressure be 6.50 atm? Rearrange the formula to solve for temperature as : \(T_2 = \frac{T_1 \times P_2}{P_1}\) Plugging in known values: \(T2 = \frac{273.15 K \times 6.50 \, atm}{11.0\, atm}\) Now, calculate the temperature and return it to Celsius: \(T_2 ≈ 128.54 K \) \(T_2^{\circ}C =128.54 K - 273.15 = -144.61^{\circ}C \) c. At what temperature would the pressure be 25.0 atm? For part (c), use the same rearranged formula: \(T_2 = \frac{T_1 \times P_2}{P_1}\) Replacing the known values: \(T_2 = \frac{273.15 K \times 25.0\,atm}{11.0\, atm}\) Now, calculate the temperature and express it in Celsius: \(T_2 ≈ \) 621.64 K \(T_2^{\circ}C = 621.64 K - 273.15 = 348.49^{\circ}C \) So the solutions are: a. Pressure at \(45^{\circ}C\) will be about 12.76 atm; b. The temperature for a pressure of 6.50 atm would be approximately \(-144.61^{\circ}C\); c. The temperature for a pressure of 25.0 atm would be approximately \(348.49^{\circ}C\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a cornerstone of understanding how gases behave under different conditions. It links the four primary variables that describe the state of a gas: pressure (P), volume (V), temperature (T), and the number of moles (n). The formula is expressed as \( PV = nRT \), where R is the universal gas constant. This law assumes that the gases are ideal, meaning they follow certain assumptions like no intermolecular forces and the volume of gas particles is negligible compared to the container.

• Pressure (P): Measured in atmospheres (atm), it is the force the gas exerts on the walls of its container.
• Volume (V): The space occupied by the gas, usually in liters.
• Temperature (T): The average kinetic energy of the gas molecules, measured in Kelvin.
• Moles (n): Amount of substance, representing the number of particles.

By applying this law, we can predict how changing one variable affects the others, helping us understand reactions involving gases and conditions like those in the problems given.

Pressure-Temperature Relationship

The pressure-temperature relationship is a crucial concept in understanding gas behavior. It arises from the observation that as temperature increases, the pressure exerted by a gas also increases, provided the volume remains constant. This relationship is depicted using Gay-Lussac's Law, which mathematically describes it as:\[\frac{P_1}{T_1} = \frac{P_2}{T_2}\]This equation shows that pressure and temperature are directly proportional when volume is unchanged. For instance, in a pressure cooker, as the heat increases, the temperature and subsequently the pressure inside rises, cooking food faster. This concept helps solve questions like those presented, where transformations from initial to final states require temperature or pressure calculations, keeping volume constant. To perform any calculation, ensure the temperature is in Kelvin to maintain consistency with the scales standard.

Kelvin Temperature Scale

The Kelvin temperature scale is essential when working with gas laws, allowing for accurate and meaningful computations. Since gas laws require absolute temperatures, Kelvin is preferred over Celsius or Fahrenheit, as it begins at absolute zero (-273.15°C), where molecular motion theoretically ceases.

• Conversion from Celsius to Kelvin: Add 273.15 to the Celsius temperature.
• Benefits of Kelvin: Avoids negative temperatures, making it beneficial for scientific calculations where proportional relationships matter.

For example, in our exercise, we converted all temperatures from Celsius to Kelvin to ensure consistent and precise application of the gas laws. Without this conversion, the direct proportionality assumption in equations like those of Gay-Lussac's Law would not hold true.

Gay-Lussac's Law

Gay-Lussac's Law is a specific instance of the gas laws focusing on the relationship between pressure and temperature, while keeping the volume constant. It asserts that the pressure of a gas is directly proportional to its temperature when measured in Kelvin. To express this, we use the formula:\[\frac{P_1}{T_1} = \frac{P_2}{T_2} \]This law simplifies the process of comparing initial and final pressure and temperature conditions in a gas-filled system. If you increase the temperature, the gas molecules move more energetically, increasing the pressure. Conversely, cooling the gas results in lower pressure.
To solve problems using Gay-Lussac's Law, such as finding the final pressure at a new temperature or determining the temperature for a desired pressure, it's crucial to:

• Ensure temperatures are in Kelvin to maintain proportionality.
• Keep volume constant as per the law's conditions.

This understanding was key in determining the changes in pressure and temperature in the provided problems, as it allowed for the straightforward application of these principles.