Question

1. calculate the root - mean - square speed of the molecules in a sample of nitrogen gas, n₂, at 25°c.

a. 612 m/s
b. 515 m/s
c. 877 m/s
d. 892 m/s
e. 933 m/s

Explanation:

Step1: Convert temperature to Kelvin

The formula for root - mean - square speed \(v_{rms}=\sqrt{\frac{3RT}{M}}\) requires temperature in Kelvin.
\(T=(25 + 273.15)\text{ K}=298.15\text{ K}\)

Step2: Determine the molar mass \(M\) of \(N_2\)

The molar mass of \(N\) is \(14.01\text{ g/mol}\). For \(N_2\), \(M = 2\times14.01\text{ g/mol}=28.02\text{ g/mol}=0.02802\text{ kg/mol}\)

Step3: Substitute values into the formula

\(R = 8.314\text{ J/(mol·K)}\)
\(v_{rms}=\sqrt{\frac{3\times8.314\times298.15}{0.02802}}\)
First, calculate the numerator: \(3\times8.314\times298.15=3\times8.314\times298.15 = 7432.7199\)
Then, \(v_{rms}=\sqrt{\frac{7432.7199}{0.02802}}=\sqrt{265265.164}\approx 515\text{ m/s}\)

Answer:

B. \(515\text{ m/s}\)