Problem 4 - University Physics with Modern Physics

Question

A \(3.00-\mathrm{L}\) tank contains air at 3.00 \(\mathrm{atm}\) and \(20.0^{\circ} \mathrm{C} .\) The tank is sealed and cooled until the pressure is 1.00 atm. (a) What is the temperature then in degrees Celsius? Assume the volume of the tank is constant. (b) If the temperature is kept at the value found in part (a) and the gas is compressed, what is the vohme when the pressure again becomes 3.00 \(\mathrm{atm} ?\)

Step-by-Step Solution

Answer

The final temperature is -175.43°C and the volume at 3.00 atm is 1.00 L.

Step 1: Understanding the Problem

We have a tank at an initial pressure of 3.00 atm and a temperature of 20°C. It is then sealed and cooled until the pressure drops to 1.00 atm. We are asked to find the final temperature in degrees Celsius. Then, while maintaining that temperature, we compress the gas back to 3.00 atm and need to find the new volume.

Step 2: Convert Initial Temperature to Kelvin

Since gas laws require temperature in Kelvin, convert the initial Celsius temperature to Kelvin. The formula is \[T(K) = T(°C) + 273.15\]. So, the initial temperature is \[T_1 = 20.0 + 273.15 = 293.15 \mathrm{K}\].

Step 3: Apply Gay-Lussac's Law for Temperature and Pressure

Gay-Lussac's Law relates pressure and temperature for a given volume: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). Given \(P_1 = 3.00 \, \mathrm{atm}\), \(T_1 = 293.15 \, \mathrm{K}\), and \(P_2 = 1.00 \, \mathrm{atm}\), solve for \(T_2\).

Step 4: Calculate Final Temperature in Kelvin

Rearrange \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\) to find \(T_2 = \frac{P_2 \times T_1}{P_1}\). Substituting, \(T_2 = \frac{1.00 \, \mathrm{atm} \times 293.15 \, \mathrm{K}}{3.00 \, \mathrm{atm}} = 97.72 \, \mathrm{K}\).

Step 5: Convert Final Temperature to Celsius

Convert the final temperature back to degrees Celsius: \[T(°C) = T(K) - 273.15\]. Thus, \[T_2(°C) = 97.72 - 273.15 = -175.43^{\circ}\mathrm{C}\].

Step 6: Apply Boyle's Law for Pressure and Volume

With constant temperature, Boyle's Law applies: \(P_1V_1 = P_2V_2\). We have \(P_2 = 1.00 \, \mathrm{atm}\), but we need to find \(V_2\) when \(P_3 = 3.00 \, \mathrm{atm}\). Rearrange to find \(V_2 = \frac{P_2V_1}{P_3}\).

Step 7: Calculate New Volume at 3.00 atm

Substituting into the equation from the previous step, \(V_2 = \frac{1.00 \, \mathrm{atm} \times 3.00 \, \mathrm{L}}{3.00 \, \mathrm{atm}} = 1.00 \, \mathrm{L}\).

Key Concepts

Gas Laws

Gas laws are a set of laws in thermodynamics that describe how gases behave under different conditions of pressure, volume, and temperature. These laws are essential for understanding how gases interact in controlled environments like a tank or container. The three main gas laws we often refer to are Boyle's Law, Charles's Law, and Gay-Lussac's Law, each dealing with a different variable relationship while keeping one variable constant.
Understanding these laws helps us predict how changes in one aspect of a gas's condition can affect another, making them incredibly useful in fields ranging from chemistry to engineering. In the case of our problem, Gay-Lussac's Law and Boyle's Law play crucial roles in understanding the behavior of the gas inside the tank.

Gay-Lussac's Law

Gay-Lussac's Law describes the relationship between pressure and temperature of a gas at constant volume. It states that the pressure of a gas is directly proportional to its absolute temperature, provided the volume remains unchanged. Mathematically, this relation is expressed as \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\).
In the original exercise, we use Gay-Lussac's Law to determine how the pressure decrease in a tank, from 3.00 atm to 1.00 atm, results in a change in temperature, assuming the volume does not change. By applying this law, we find the new temperature when the pressure is different, helping us understand the cooling process in the tank.

Boyle's Law

Boyle's Law explains the inverse relationship between the pressure and volume of a gas at constant temperature. This means that when the pressure of a gas increases, its volume decreases proportionally, as long as the temperature remains the same. The mathematical expression for this relationship is \(P_1V_1 = P_2V_2\).
In our exercise, Boyle's Law helps us understand what happens when we compress the gas in the tank back to the original pressure of 3.00 atm, after cooling it. Since the temperature is kept constant at this point, we can calculate the new reduced volume owing to the pressure change, using the formula to find that the volume decreases to 1.00 L.

Pressure-Volume Relationship

The pressure-volume relationship is central to the gas laws, particularly Boyle's Law. It describes how the volume of a gas changes in response to changes in pressure when temperature is held steady. This relationship is vital for predicting how gases behave under compression.
In practical terms, understanding this relationship allows scientists and engineers to design equipment and processes that can safely and efficiently handle gases under varying pressures. In the exercise, this concept helped to find out the new volume of gas when the pressure was returned to its initial state, demonstrating the interplay between these two key variables. This understanding is crucial for applications involving gas storage and transfer.