To calculate the fraction of $\text{N}_2$ molecules that have speeds in the range of 260 to 270 m/s at a temperature of 280 K, we use the Maxwell-Boltzmann distribution formula for the speed distribution of gas molecules:

$f(v) = 4\pi \left( \frac{m}{2\pi k_B T} \right)^{3/2} v^2 \exp\left(-\frac{mv^2}{2k_B T}\right)$

Where:

• $f(v)$ is the probability density function for speed $v$,
• $m$ is the mass of a nitrogen molecule $\text{N}_2$,
• $k_B$ is the Boltzmann constant,
• $T$ is the temperature in Kelvin,
• $v$ is the speed of the molecule.

First, we calculate the necessary constants:

• Molar mass of $\text{N}_2$ = 28.014 g/mol
• $m = \frac{28.014 \times 10^{-3}}{6.022 \times 10^{23}}$ kg (mass of one molecule)
• $k_B = 1.38 \times 10^{-23}$ J/K

Next, integrate the Maxwell-Boltzmann distribution over the given range [260 m/s, 270 m/s] to find the fraction of molecules with speeds in that range:

$\text{Fraction} = \int_{260}^{270} f(v) \, dv$

I'll calculate the exact value now.The fraction of $\text{N}_2$ molecules that have speeds in the range of 260 to 270 m/s at a temperature of 280 K is approximately 0.015.

Would you like further details or have any questions? Here are some related questions that might interest you:

1. How does the temperature affect the Maxwell-Boltzmann speed distribution?
2. What is the most probable speed for nitrogen molecules at 280 K?
3. How would the fraction change if the temperature were increased to 300 K?
4. What fraction of molecules have speeds between 270 m/s and 280 m/s?
5. How does the Maxwell-Boltzmann distribution differ for gases of different molar masses?

Tip: The Maxwell-Boltzmann distribution provides insight into how molecular speeds vary with temperature, which is essential in understanding gas behavior in various thermodynamic processes.