Find Vrms for a nitrogen gas N₂ at 33°C. 4. 622.0 m/s 479.0 m/s 462.4 m/s ABC A. B. C. 1--1 7 N 14.01 D. E. F. 522.1 m/s 626.5 m/s 598.3 m/s

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### Root Mean Square Speed of Nitrogen Gas

**Problem Statement:**
Find the root mean square speed (\(v_{\text{rms}}\)) for nitrogen gas (\(N_2\)) at 333°C.

**Multiple Choice Options:**
A. 6220 m/s  
B. 4790 m/s  
C. 4624 m/s  
D. 5221 m/s  
E. 6265 m/s  
F. 5083 m/s  

**Additional Problem for Context:**
If 3.8 moles of an ideal gas is at a temperature of 65°C and contained in a 0.127 cubic meter (m³) container, what is the pressure on the walls of the container due to the gas? (Use 8.314 J/(mol·K))

*Note: The periodic table entry for nitrogen is included, indicating its atomic number (7) and atomic mass (14.01 g/mol). \(N_2\) refers to the diatomic molecule of nitrogen.*

**Explanation:**
For calculating the \(v_{\text{rms}}\) of a gas, the formula to use is:
\[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \]

Where:
- \(k\) is the Boltzmann constant (\(1.38 \times 10^{-23} \text{J/K}\))
- \(T\) is the absolute temperature in Kelvin
- \(m\) is the mass of one molecule of the gas in kilograms

First, convert the given temperature from Celsius to Kelvin:
\[ T(\text{K}) = 333 + 273.15 = 606.15 \text{ K} \]

Given that the molar mass of nitrogen \(N_2\) is \(2 \times 14.01 \text{ g/mol} = 28.02 \text{ g/mol}\):

Convert the mass from grams to kilograms:
\[ m = \frac{28.02}{1000} \text{ kg/mol} \]

Finally, use the given options to identify the correct answer:
This specific problem would typically involve calculating values step by step which can be done using appropriate formulas and constants for ideal gas law and kinetic theory of gases.

In this representation, students are expected to use the formula and constants to solve the problem provided the choices A-F, checking the consistency with the
Transcribed Image Text:### Root Mean Square Speed of Nitrogen Gas **Problem Statement:** Find the root mean square speed (\(v_{\text{rms}}\)) for nitrogen gas (\(N_2\)) at 333°C. **Multiple Choice Options:** A. 6220 m/s B. 4790 m/s C. 4624 m/s D. 5221 m/s E. 6265 m/s F. 5083 m/s **Additional Problem for Context:** If 3.8 moles of an ideal gas is at a temperature of 65°C and contained in a 0.127 cubic meter (m³) container, what is the pressure on the walls of the container due to the gas? (Use 8.314 J/(mol·K)) *Note: The periodic table entry for nitrogen is included, indicating its atomic number (7) and atomic mass (14.01 g/mol). \(N_2\) refers to the diatomic molecule of nitrogen.* **Explanation:** For calculating the \(v_{\text{rms}}\) of a gas, the formula to use is: \[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \] Where: - \(k\) is the Boltzmann constant (\(1.38 \times 10^{-23} \text{J/K}\)) - \(T\) is the absolute temperature in Kelvin - \(m\) is the mass of one molecule of the gas in kilograms First, convert the given temperature from Celsius to Kelvin: \[ T(\text{K}) = 333 + 273.15 = 606.15 \text{ K} \] Given that the molar mass of nitrogen \(N_2\) is \(2 \times 14.01 \text{ g/mol} = 28.02 \text{ g/mol}\): Convert the mass from grams to kilograms: \[ m = \frac{28.02}{1000} \text{ kg/mol} \] Finally, use the given options to identify the correct answer: This specific problem would typically involve calculating values step by step which can be done using appropriate formulas and constants for ideal gas law and kinetic theory of gases. In this representation, students are expected to use the formula and constants to solve the problem provided the choices A-F, checking the consistency with the
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