Calculate the root mean square velocities of \(\mathrm{CH}_{4}(g)\) and \(\mathrm{N}_{2}(g)\) molecules at 273 \(\mathrm{K}\) and 546 \(\mathrm{K} .\)

Short Answer

Expert verified
The root mean square velocities for CH4 and N2 at the given temperatures are: For CH4 at 273 K: \(v_{rms} \approx 922.65\,\mathrm{m/s}\) For CH4 at 546 K: \(v_{rms} \approx 1304.97\,\mathrm{m/s}\) For N2 at 273 K: \(v_{rms} \approx 661.54\,\mathrm{m/s}\) For N2 at 546 K: \(v_{rms} \approx 935.68\,\mathrm{m/s}\)

Step by step solution

01

Write down the equation for the root mean square (rms) velocity

To find the root mean square velocity of a gas, we will use the following formula: \[v_{rms} = \sqrt{\frac{3RT}{M}}\] Where \(v_{rms}\) is the root mean square velocity, \(R\) is the ideal gas constant (8.314 J/mol K), \(T\) is the temperature in Kelvin, and \(M\) is the molecular mass in kg/mol.
02

Calculate the molecular mass of CH4 and N2

To find the molecular mass of CH4 and N2 in kg/mol, we'll need to multiply the atomic mass of each element by its frequency in the molecule and then convert the result to kg. For CH4: \(M_{CH4} = (12.01\,\mathrm{g/mol}\times 1) + (1.01\,\mathrm{g/mol}\times 4)\) \(M_{CH4} = 12.01\,\mathrm{g/mol} + 4.04\,\mathrm{g/mol}\) \(M_{CH4} = 16.05\,\mathrm{g/mol}\) Converting to kg/mol, we get: \(M_{CH4} = 0.01605\,\mathrm{kg/mol}\) For N2: \(M_{N2}= (14.01\,\mathrm{g/mol}\times 2)\) \(M_{N2} = 28.02\,\mathrm{g/mol}\) Converting to kg/mol, we get: \(M_{N2} = 0.02802\,\mathrm{kg/mol}\)
03

Calculate the root mean square velocities

Now that we have the molecular masses and the temperature values, we can calculate the root mean square velocities for CH4 and N2 at 273 K and 546 K. For CH4 at 273 K: \(v_{rms} = \sqrt{\frac{3\times 8.314\,\mathrm{J/(mol\cdot K)}\times 273\,\mathrm{K}}{0.01605\,\mathrm{kg/mol}}}\) \(v_{rms} \approx 922.65\,\mathrm{m/s}\) For CH4 at 546 K: \(v_{rms} = \sqrt{\frac{3\times 8.314\,\mathrm{J/(mol\cdot K)}\times 546\,\mathrm{K}}{0.01605\,\mathrm{kg/mol}}}\) \(v_{rms} \approx 1304.97\,\mathrm{m/s}\) For N2 at 273 K: \(v_{rms} = \sqrt{\frac{3\times 8.314\,\mathrm{J/(mol\cdot K)}\times 273\,\mathrm{K}}{0.02802\,\mathrm{kg/mol}}}\) \(v_{rms} \approx 661.54\,\mathrm{m/s}\) For N2 at 546 K: \(v_{rms} = \sqrt{\frac{3\times 8.314\,\mathrm{J/(mol\cdot K)}\times 546\,\mathrm{K}}{0.02802\,\mathrm{kg/mol}}}\) \(v_{rms} \approx 935.68\,\mathrm{m/s}\)
04

Present the final results

The root mean square velocities for each gas at the given temperatures are: For CH4 at 273 K: \(v_{rms} \approx 922.65\,\mathrm{m/s}\) For CH4 at 546 K: \(v_{rms} \approx 1304.97\,\mathrm{m/s}\) For N2 at 273 K: \(v_{rms} \approx 661.54\,\mathrm{m/s}\) For N2 at 546 K: \(v_{rms} \approx 935.68\,\mathrm{m/s}\)

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