Problem 4 - University Physics with Modern Physics

Question

A \(3.00-\) L tank contains air at 3.00 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 that 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 volume when the pressure again becomes 3.00 atm?

Step-by-Step Solution

Answer

(a) -175.43°C temperature; (b) 1.00 L volume.

Step 1: Understand the Problem

We have a tank with a constant volume of 3.00 L. The initial conditions are an air pressure of 3.00 atm and a temperature of 20.0°C. The problem requires us to find the final temperature when the tank's pressure is reduced to 1.00 atm, assuming the volume doesn't change.

Step 2: Convert Initial Temperature

Convert the initial temperature from Celsius to Kelvin by using the formula: \( T(K) = T(°C) + 273.15 \). Therefore, \( T_1 = 20.0 + 273.15 = 293.15 \, K \).

Step 3: Apply Ideal Gas Law

Since the volume is constant, use the formula \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \). Solve for \( T_2 \): \( T_2 = \frac{P_2 \times T_1}{P_1} \). Substitute \( P_1 = 3.00 \) atm, \( T_1 = 293.15 \) K, and \( P_2 = 1.00 \) atm.

Step 4: Calculate Final Temperature in Kelvin

Plug the values into the equation: \( T_2 = \frac{1.00 \times 293.15}{3.00} = 97.72 \) K.

Step 5: Convert Final Temperature to Celsius

Convert the temperature back to Celsius: \( T(°C) = T(K) - 273.15 \). Thus, \( T_2 = 97.72 - 273.15 = -175.43 \, °C \).

Step 6: Understand the Second Problem Part

Now, an isothermal compression is considered where the final temperature remains the same as in part (a), \(-175.43 \, °C\). We want to find the new volume when the pressure reaches 3.00 atm again.

Step 7: Apply Boyle's Law

During an isothermal process, use Boyle's Law: \( P_1 V_1 = P_2 V_2 \). Solve for \( V_2 \): \( V_2 = \frac{P_1 V_1}{P_2} \). Substitute \( P_1 = 1.00 \) atm, \( V_1 = 3.00 \) L, and \( P_2 = 3.00 \) atm.

Step 8: Calculate Final Volume

Substitute into the equation: \( V_2 = \frac{1.00 \times 3.00}{3.00} = 1.00 \, L \).

Key Concepts

Boyle's Law

Boyle's Law is a fundamental principle in understanding gas behavior under pressure. It describes the relationship between the pressure and volume of a gas in an isothermal process, where the temperature remains constant. According to Boyle's Law, the product of the pressure and volume of a gas is a constant for a given mass of gas at a constant temperature. This relationship is mathematically expressed as:\[ P_1 V_1 = P_2 V_2 \]Here:- \(P_1\) and \(P_2\) are the initial and final pressures, respectively.- \(V_1\) and \(V_2\) are the initial and final volumes, respectively.The law indicates that if the volume of a gas decreases, its pressure increases if the temperature remains unchanged, and vice versa. Boyle's Law is particularly useful when studying processes where heat is not exchanged with the surroundings, such as adiabatic processes in insulated environments. Remember, this law only holds when temperature remains constant throughout the process.

temperature conversion

Temperature conversion is a crucial step in dealing with gas law problems. Temperature in the Celsius scale is commonly used; however, in gas laws, it must be converted to Kelvin because Kelvin is the absolute temperature scale.To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature:- **Conversion Formula:** \( T(K) = T(^\circ\text{C}) + 273.15 \)For example, to convert \(20.0 \, ^\circ\text{C}\) to Kelvin:- \( T = 20.0 + 273.15 = 293.15 \, \text{K} \)Kelvin is used because it allows all the gas laws' equations to work without negative temperatures, as absolute zero (0 K) is the lowest possible temperature, where no thermal energy remains. This ensures precise calculations and avoids complications associated with negative values when applying gas laws.

isothermal process

An isothermal process in thermodynamics is a process in which the temperature of the system remains constant, while changes in pressure and volume occur. For such a process, any heat added to the system is used to perform work, maintaining a constant temperature.Isothermal processes can be represented by the equation for Boyle's Law, since the temperature is constant, establishing a direct relationship between pressure and volume:\[ P_1 V_1 = P_2 V_2 \]In real-world applications, isothermal processes can occur when a gas is contained in a sealed system that allows for slow equilibrium with its surroundings, ensuring no temperature change as the gas expands or compresses.Examples include:

• Compression or expansion of gases in engines.
• Piston mechanisms where the system is allowed to reach thermal equilibrium steadily.

These processes are idealized scenarios, often requiring specific conditions to be perfect, like sufficient time for heat exchange, but they are crucial for understanding basic thermodynamic principles.