Problem 34 - University Physics with Modern Physics

Question

A container with volume 1.48 \(\mathrm{L}\) is initially evacuated. Then it is filled with 0.226 \(\mathrm{g}\) of \(\mathrm{N}_{2} .\) Assume that the pressure of the gas is low enough for the gas to obey the ideal-gas law to high degree of accuracy. If the root-mean-square speed of the gas molecules is \(182 \mathrm{m} / \mathrm{s},\) what is the pressure of the gas?

Step-by-Step Solution

Answer

The pressure of the gas is approximately 134.45 Pa.

Step 1: Understand the Relationship

The root-mean-square speed of gas molecules is given by the equation \[ v_{rms} = \sqrt{\frac{3RT}{M}} \]where \( v_{rms} \) is the root-mean-square speed, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas. We will use this to find the temperature \( T \) first.

Step 2: Convert Volume to Cubic Meters

Convert the given volume from liters to cubic meters because SI units are preferred for consistency in gas law calculations.1.48 L is equivalent to \( 1.48 \times 10^{-3} \) m³.

Step 3: Find Molar Mass of Nitrogen

The molar mass \( M \) of nitrogen \( \mathrm{N}_{2} \) is calculated as 28.02 g/mol. Convert it to kilograms per mole (the SI unit) by dividing by 1000:\[ M = 0.02802 \text{ kg/mol} \].

Step 4: Calculate Temperature Using Root-Mean-Square Speed

Using the root-mean-square formula, solve for \( T \):\[ 182 = \sqrt{\frac{3RT}{0.02802}} \]Squaring both sides and solving for \( T \), we get:\[ 182^2 = \frac{3 \cdot 8.314 \cdot T}{0.02802} \]\[ T = \frac{0.02802 \cdot 182^2}{3 \cdot 8.314} \approx 29.51 \text{ K} \].

Step 5: Use Ideal Gas Law to Find Pressure

Now using the ideal gas law:\[ PV = nRT \]First, calculate \( n \), the number of moles, as \( \frac{\text{mass}}{\text{molar mass}} \):\[ n = \frac{0.226}{28.02 \times 10^{-3}} \approx 0.00806 \text{ mol} \].Substitute \( n, R, T, \) and \( V \) into the ideal gas law:\[ P \times 1.48 \times 10^{-3} = 0.00806 \times 8.314 \times 29.51 \]Solve for \( P \):\[ P = \frac{0.00806 \times 8.314 \times 29.51}{1.48 \times 10^{-3}} \approx 134.45 \text{ Pa} \].

Key Concepts

Root-Mean-Square Speed

Understanding the root-mean-square speed of a gas is essential in thermodynamics. It's a measure that reflects the speed at which gas molecules move, averaged over time. The formula is given by: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \]In this equation, \( v_{rms} \) stands for the root-mean-square speed, \( R \) is the ideal gas constant \( (8.314 \, \mathrm{J/mol \, K}) \), \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas. The higher the \( v_{rms} \), the faster the molecules are moving on average. Thus, knowing the \( v_{rms} \) aids in calculating other variables like temperature when pressure or volume is not available. It essentially offers a window into the kinetic energy of the molecules within the system. Since kinetic energy and temperature are related, \( v_{rms} \) provides valuable insight into the thermal state of the gas.

Molar Mass

Molar mass is a fundamental concept in chemistry and is crucial when dealing with gas laws. It represents the mass of one mole of a substance. In the case of nitrogen \( \mathrm{N}_2 \), the molar mass is approximately \( 28.02 \, \mathrm{g/mol} \). When using the root-mean-square formula, it is often necessary to convert this to kilograms per mole, thus: \[ M = \frac{28.02}{1000} = 0.02802 \, \mathrm{kg/mol} \]. Using kilograms per mole is convenient for calculations involving the ideal gas law, which requires SI units. Understanding molar mass allows scientists to transition from understanding microscopic properties to macroscopic properties, like calculating the number of moles from a given mass. It serves as a bridge between the mass of individual molecules and the bulk properties of gases, which is pivotal when using equations like the ideal gas law.

Temperature Calculation

Temperature is a crucial variable in gas law equations. In the context of this exercise, we calculate temperature from the root-mean-square speed using the formula: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \]. By squaring both sides and solving for \( T \), we can re-arrange the formula as:\[ T = \frac{M \cdot v_{rms}^2}{3R} \]. Here, the calculated temperature was approximately \( 29.51 \, \mathrm{K} \). It illustrates the direct relationship between molecular speed and temperature. As the temperature of a gas increases, its molecules move faster, increasing the \( v_{rms} \). This relationship is essential for predicting how gases will behave under changing conditions. It reflects the kinetic theory of gases, which posits that temperature is proportional to average molecular kinetic energy. Understanding this concept is vital for applying the ideal gas law to real-world problems.