Problem 9 - University Physics with Modern Physics

Question

A large cylindrical tank contains 0.750 in \(^{3}\) of nitrogen gas at \(27^{\circ} \mathrm{C}\) and \(1.50 \times 10^{5} \mathrm{Pa}\) (absolute pressure). The tank has a tight-fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to 0.480 \(\mathrm{m}^{3}\) and the temperature is increased to \(157^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

The final pressure is approximately \(3.36 \times 10^5\) Pa.

Step 1: Define the Ideal Gas Law

The ideal gas law is given by the formula \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the amount of substance in moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

Step 2: Convert Temperatures to Kelvin

The initial temperature \( T_1 \) in Celsius is 27°C, which is equal to \( T_1 = 27 + 273.15 = 300.15 \) K. The final temperature \( T_2 \) is 157°C, which is equal to \( T_2 = 157 + 273.15 = 430.15 \) K.

Step 3: Apply the Combined Gas Law

Since \( n \) and \( R \) are constant, we use \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) to find \( P_2 \). Here, \( P_1 = 1.50 \times 10^5 \) Pa, \( V_1 = 0.750 \) m³, and \( T_1 = 300.15 \) K. \( V_2 = 0.480 \) m³ and \( T_2 = 430.15 \) K.

Step 4: Solve for Final Pressure \( P_2 \)

Rearrange the equation to solve for \( P_2 \): \[ P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} \]. Substitute the values: \( P_2 = 1.50 \times 10^5 \times \frac{0.750}{0.480} \times \frac{430.15}{300.15} \).

Step 5: Calculate the Value of \( P_2 \)

Calculate the expression: \( P_2 = 1.50 \times 10^5 \times 1.5625 \times 1.433 \). Compute to get \( P_2 \approx 3.36 \times 10^5 \) Pa.

Key Concepts

Gas Laws

The key concept behind the given exercise is the Ideal Gas Law, which is a cornerstone of gas behavior studies. This law combines several other gas laws and provides a generalized equation to describe the state of an ideal gas. The law is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is the volume occupied by the gas, \( n \) represents the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

Explaining each variable helps to understand their role:

• **Pressure (P)**: This is the force exerted by the gas particles against the walls of the container per unit area.
• **Volume (V)**: Represents the space the gas occupies.
• **Moles (n)**: A measure of the quantity of gas particles present.
• **Temperature (T)**: Impacts the speed and energy of the gas particles, which in turn affects pressure and volume.

Understanding these variables and their interrelations is crucial for solving problems related to changing states of gases. For cases where the amount of gas and the constant \( R \) remain the same, the Combined Gas Law \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \) is often used to solve problems as in this exercise, enabling the calculation of a new pressure after changes in volume or temperature.

Thermodynamics

Thermodynamics is the branch of physics concerned with heat and temperature and their relation to energy and work. It incorporates how gas laws integrate into broader thermodynamic concepts, explaining how energy is transferred in the form of heat and work, which affects pressure and volume changes in gases.

In this exercise, the temperature of the gas initially at \( 27^{\circ} \mathrm{C} \) increases to \( 157^{\circ} \mathrm{C} \). To facilitate calculations using the Ideal Gas Law, temperature is converted into Kelvin, which sets the absolute scale avoiding negative values that are non-physical in this context. The Kelvin scale is directly related to energy; higher Kelvin numbers indicate more kinetic energy, impacting gas behavior.

Thermodynamic processes influencing gas laws can be described through terms like **isothermal**, where temperature remains constant, or **adiabatic**, where no heat exchange occurs. In this case, it's neither, as both temperature and volume change, adhering to **non-isothermal** conditions. Understanding these concepts helps in predicting outcomes when modifying temperature and volume in gas systems.

Pressure Calculations

Pressure calculation is a focal theme in gas laws. In this problem, the **final pressure** of the gas is determined by changes in both volume and temperature. The formula used, derived from the Combined Gas Law, simplifies to \[ P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} \].

Each component of this equation represents a factor contributing to the new pressure:

• **Initial Pressure \( (P_1) \)**: Acts as a starting point for calculations.
• **Volume Ratio \( (\frac{V_1}{V_2}) \)**: A smaller final volume \( V_2 \) compared to the initial volume \( V_1 \) will increase pressure, assuming constant temperature initially under Boyle’s Law expectations.
• **Temperature Ratio \( (\frac{T_2}{T_1}) \)**: A higher final temperature \( T_2 \), compared to \( T_1 \), increases pressure as gases expand according to Charles’s Law.

Thus, converting and combining these ratios leads to the calculated final pressure \( P_2 \), showing how these physical changes impact gas behavior in a quantifiable manner. Understanding these calculations is essential for predicting and controlling gas systems in practical applications, like in the case of pistons within engines or measuring atmospheric pressure changes.