Question

A container is filled with an ideal gas to a pressure of \(40.0\) 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 \(1.50 \times 10^{2}\) atm? c. At what temperature would the pressure be \(25.0 \mathrm{~atm} ?\)

Short answer

a. The pressure in the container at \(45^{\circ} \mathrm{C}\) is approximately \(46.6 \mathrm{~atm}\). b. The temperature at which the pressure would be \(1.50 \times 10^{2} \mathrm{~atm}\) is approximately \(1024.73 \mathrm{~K}\). c. The temperature at which the pressure would be \(25.0 \mathrm{~atm}\) is approximately \(171.22 \mathrm{~K}\).

Step-by-step solution

Convert temperatures to Kelvin

First, we need to convert temperatures given in Celsius to Kelvin. We can do this by adding 273.15 to the Celsius temperature. T1 = 0°C + 273.15 = 273.15 K T2 = 45°C + 273.15 = 318.15 K

Calculate the pressure at 45°C (318.15 K)

Use the pressure-temperature relationship formula and the given initial pressure and temperatures to calculate the pressure at the final temperature of 45°C (318.15 K). The formula for the pressure-temperature relationship is: (P1/T1) = (P2/T2) Plugging in the values given in the exercise: (40.0 atm / 273.15 K) = (P2 / 318.15 K) Solve for P2: P2 = (40.0 atm) * (318.15 K / 273.15 K) P2 ≈ 46.6 atm a. The pressure in the container at 45°C is approximately 46.6 atm.

Calculate the temperature at 150 atm

Now, we need to calculate the temperature at which the pressure is equal to 150 atm. Using the pressure-temperature relationship formula: (P1/T1) = (P2/T2) Plugging in the values given in the exercise and the new pressure: (40.0 atm / 273.15 K) = (150 atm / T2) Solve for T2: T2 = (150 atm) * (273.15 K / 40.0 atm) T2 ≈ 1024.725 K b. The temperature at which the pressure would be 150 atm is approximately 1024.73 K.

Calculate the temperature at 25 atm

Lastly, we will calculate the temperature at which the pressure is equal to 25 atm. Using the pressure-temperature relationship formula: (P1/T1) = (P2/T2) Plugging in the values given in the exercise and the new pressure: (40.0 atm / 273.15 K) = (25 atm / T2) Solve for T2: T2 = (25 atm) * (273.15 K / 40.0 atm) T2 ≈ 171.22 K c. The temperature at which the pressure would be 25 atm is approximately 171.22 K.

Key concepts

Pressure-Temperature Relationship

The pressure-temperature relationship is a fundamental concept in gas laws, particularly the Ideal Gas Law. This relationship is expressed through the equation \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), where \( P \) represents pressure and \( T \) represents temperature in Kelvin.

For ideal gases, this relationship shows that the pressure is directly proportional to the temperature, as long as the volume and the amount of gas remain constant. When the temperature of a gas increases, so does its pressure, provided the gas does not expand or contract.

**Key Points**

• An increase in temperature raises the kinetic energy of gas particles, causing more frequent and forceful collisions with the container walls, thereby increasing pressure.
• This relationship helps predict how a change in temperature affects pressure, which is crucial in many scientific and engineering calculations.

Kelvin Temperature Scale

The Kelvin temperature scale is an absolute scale used in scientific calculations. It starts at absolute zero, the lowest possible temperature where molecular motion ceases. Unlike Celsius or Fahrenheit, Kelvin does not use degrees as a unit.

**Why Kelvin?**

• Kelvin provides a direct correlation with energy; temperatures measured in Kelvin can be seamlessly used in equations involving physical laws such as the Ideal Gas Law.
• The use of Kelvin eliminates negative numbers, making mathematical manipulation more straightforward.

To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. For example, \(0^{\circ} \mathrm{C}\) becomes \(273.15 \mathrm{K}\). Using Kelvin ensures consistency in thermodynamic calculations and allows for the direct application of relationships like \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \).

Pressure Calculation

Pressure calculation in gases involves understanding how changes in temperature, under constant volume, affect the gas's pressure. To calculate new pressure conditions based on changes in temperature, you can rearrange the pressure-temperature relationship formula.

For instance, if you start with an initial pressure \( P_1 \) at temperature \( T_1 \), and want to find the new pressure \( P_2 \) at a temperature \( T_2 \):
1. Use the equation \( P_2 = P_1 \times \frac{T_2}{T_1} \)
2. Plug in the values and solve for \( P_2 \).

**Example Application**

• If you have a gas at \(40 \; \mathrm{atm}\) and \(0^{\circ} \mathrm{C}\), and heat it to \(45^{\circ} \mathrm{C}\), convert these to Kelvin and apply:
  \( T_1 = 273.15 \; \mathrm{K}\), \( T_2 = 318.15 \; \mathrm{K}\). Calculate \( P_2 \), finding it to be approximately \(46.6 \; \mathrm{atm}\).

This way, pressure calculation helps predict and analyze the behavior of gases in changing thermal environments.