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

a. The pressure after heating the container to \(45^{\circ} C\) will be approximately \(12.88 \text{ atm}\). b. The temperature at which the pressure would be \(6.50 \text{ atm}\) is approximately \(152.99 \text{ K}\). c. The temperature at which the pressure would be \(25.0 \text{ atm}\) is approximately \(618.75 \text{ K}\).

Step-by-step solution

Convert the given temperatures to Kelvin

To convert a temperature from Celsius to Kelvin, we add 273.15 to the Celsius temperature. The initial temperature (\(T_1\)) in Celsius is \(0^{\circ} C\), so the initial temperature in Kelvin is: \(T_1\) = \(0 + 273.15\) = \(273.15 K\) The final temperature (\(T_2\)) in Celsius is \(45^{\circ} C\), so the final temperature in Kelvin is: \(T_2\) = \(45 + 273.15\) = \(318.15 K\)

Apply Gay-Lussac's Law formula to find the final pressure

We know the initial pressure (\(P_1\)) and the initial and final temperatures (\(T_1\) and \(T_2\)). We will now plug these values into Gay-Lussac's Law formula and solve for the final pressure, \(P_2\). \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\) Now, rearrange the formula to solve for \(P_2\): \(P_2 = \frac{P_1 * T_2}{T_1}\) Plug in the known values and solve for \(P_2\): \(P_2 = \frac{11.0 \text{ atm} * 318.15 \text{ K}}{273.15 \text{ K}}\) \(P_2 = 12.88 \text{ atm}\) So, the pressure after heating the container to \(45^{\circ} C\) will be approximately \(12.88 \text{ atm}\). ###Part b: Find the temperature at which the pressure is 6.50 atm###

Assign known values

For this part of the problem, we are given the final pressure (\(P_2 = 6.50 \text{ atm}\)) and we already know the initial pressure and temperature (\(P_1 = 11.0 \text{ atm}\), \(T_1 = 273.15 \text{ K}\)). We will now plug these values into the Gay-Lussac's Law formula to find the temperature (\(T_2\)).

Apply Gay-Lussac's Law to find the final temperature

Rearrange the formula to solve for \(T_2\): \(T_2 = \frac{P_2 * T_1}{P_1}\) Plug in the known values and solve for \(T_2\): \(T_2 = \frac{6.50 \text{ atm} * 273.15 \text{ K}}{11.0 \text{ atm}}\) \(T_2 = 152.99 \text{ K}\) So, the temperature at which the pressure would be \(6.50 \text{ atm}\) is approximately \(152.99 \text{ K}\). ###Part c: Find the temperature at which the pressure is 25.0 atm###

Assign known values

For this part, we are given the final pressure (\(P_2 = 25.0 \text{ atm}\)) and we already know the initial pressure and temperature. We will now plug these values into the Gay-Lussac's Law formula to find the temperature (\(T_2\)).

Apply Gay-Lussac's Law to find the final temperature

Use the same rearranged formula as before: \(T_2 = \frac{P_2 * T_1}{P_1}\) Plug in the known values and solve for \(T_2\): \(T_2 = \frac{25.0 \text{ atm} * 273.15 \text{ K}}{11.0 \text{ atm}}\) \(T_2 = 618.75 \text{ K}\) So, the temperature at which the pressure would be \(25.0 \text{ atm}\) is approximately \(618.75 \text{ K}\).

Key concepts

Ideal Gas Behavior

The behavior of an ideal gas is a foundational concept in chemistry and physics. It's described by several gas laws, one of which is Gay-Lussac's Law. An ideal gas is a hypothetical gas that perfectly follows these laws under all conditions of temperature and pressure. This assumption allows for simplification in calculations and a greater understanding of gas behavior.

For an ideal gas, we assume that the particles have no volume and no intermolecular forces—that is, they do not attract or repel each other. Also, when particles collide, whether with each other or the walls of their container, these collisions are perfectly elastic, meaning no energy is lost. Real gases approximate the behavior of ideal gases at high temperatures and low pressures, as under these conditions, the forces between molecules become negligible, and their actual volume is far less significant compared to the space they occupy.

Temperature-Pressure Relationship

Gay-Lussac's Law reveals the direct relationship between the pressure and temperature of a gas, assuming the volume and the amount of gas remain constant. As the kinetic energy of a gas increases with temperature, so does the pressure, as the gas particles hit the walls of their container more frequently and with more force. This law is expressed mathematically as \( P_1/T_1 = P_2/T_2 \) 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, respectively.

The relationship is such that if the temperature of the gas increases, the pressure also increases, as long as the mass and volume of the gas do not change. Conversely, if the temperature decreases, the pressure likewise decreases. This behavior is crucial for many practical applications such as in predicting the behavior of gases within various temperature ranges.

Converting Celsius to Kelvin

Understanding temperature scales is key to working with gas laws, as the laws are designed to use the Kelvin scale. The Kelvin (K) is the SI unit for temperature and is used in scientific measurements because it is an absolute scale, starting at absolute zero, the point at which particles theoretically stop moving.

To convert a temperature from Celsius to Kelvin, which is essential before applying Gay-Lussac's Law, you add 273.15 to the Celsius temperature. This allows you to align with the absolute scale needed for the calculations. For instance, the room temperature of around \(20^\circ C\) is equivalent to \(293.15 K\). This step is imperative because using Celsius instead of Kelvin could result in inaccurate calculations and a misunderstanding of a gas's behavior under varying temperatures.