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Chapter 8: Problem 103

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 of CH4 and N2 at 273 K and 546 K are as follows: CH4 at 273 K: \(606.62 \mathrm{m/s}\), N2 at 273 K: \(455.57 \mathrm{m/s}\), CH4 at 546 K: \(858.80 \mathrm{m/s}\), and N2 at 546 K: \(644.60 \mathrm{m/s}\).

Step by step solution

01

Calculate molecular masses of CH4 and N2

To calculate the molecular masses, we need to use the atomic masses of carbon (C), hydrogen (H), and nitrogen (N). Atomic masses (in atomic mass units - amu): - Carbon (C): 12.01 amu - Hydrogen (H): 1.01 amu - Nitrogen (N): 14.01 amu Molecular mass of CH4: 1 C atom + 4 H atoms = (1 x 12.01) + (4 x 1.01) = 12.01 + 4.04 = 16.05 amu Molecular mass of N2: 2 N atoms = 2 x 14.01 = 28.02 amu To convert these molecular masses to kilograms, we need to multiply by the conversion factor: 1 amu = \(1.66 \times 10^{-27} \mathrm{kg}\) Mass of CH4: 16.05 amu x \(1.66 \times 10^{-27} \mathrm{kg/amu}\) = \(2.66 \times 10^{-26} \mathrm{kg}\) Mass of N2: 28.02 amu x \(1.66 \times 10^{-27} \mathrm{kg/amu}\) = \(4.65 \times 10^{-26} \mathrm{kg}\)
02

Calculate root mean square velocities at 273 K

Now we will use the root mean square velocity formula to calculate the velocities at 273 K: CH4 at 273 K: \(v_{rms} = \sqrt{\frac{3(1.38 \times 10^{-23})(273)}{2.66 \times 10^{-26}}}\) = \(606.62 \mathrm{m/s}\) N2 at 273 K: \(v_{rms} = \sqrt{\frac{3(1.38 \times 10^{-23})(273)}{4.65 \times 10^{-26}}}\) = \(455.57 \mathrm{m/s}\)
03

Calculate root mean square velocities at 546 K

Now we will calculate the velocities at 546 K: CH4 at 546 K: \(v_{rms} = \sqrt{\frac{3(1.38 \times 10^{-23})(546)}{2.66 \times 10^{-26}}}\) = \(858.80 \mathrm{m/s}\) N2 at 546 K: \(v_{rms} = \sqrt{\frac{3(1.38 \times 10^{-23})(546)}{4.65 \times 10^{-26}}}\) = \(644.60 \mathrm{m/s}\)
04

Present the results

The root mean square velocities of CH4 and N2 are: - CH4 at 273 K: \(606.62 \mathrm{m/s}\) - N2 at 273 K: \(455.57 \mathrm{m/s}\) - CH4 at 546 K: \(858.80 \mathrm{m/s}\) - N2 at 546 K: \(644.60 \mathrm{m/s}\)

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Most popular questions from this chapter

From the values in Table \(8-3\) for the van der Waals constant \(a\) for the gases \(\mathrm{H}_{2}, \mathrm{CO}_{2}, \mathrm{N}_{2},\) and \(\mathrm{CH}_{4},\) predict which of these gas molecules show the strongest intermolecular attractions.

Urea \(\left(\mathrm{H}_{2} \mathrm{NCONH}_{2}\right)\) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide: $$2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{NCONH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)$$ Ammonia gas at \(223^{\circ} \mathrm{C}\) and \(90 .\) atm flows into a reactor at a rate of \(500 .\) L/min. Carbon dioxide at \(223^{\circ} \mathrm{C}\) and \(45\) atm flows into the reactor at a rate of \(600 .\) L/min. What mass of urea is produced per minute by this reaction assuming \(100 \%\) yield?

Freon-12 \(\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right)\) is commonly used as the refrigerant in central home air conditioners. The system is initially charged to a pressure of 4.8 atm. Express this pressure in each of the following units ( 1 atm \(=14.7\) psi). a. \(\mathrm{mm} \mathrm{Hg}\) b. \(torr\) c. \(Pa\) d. \(psi\)

A compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\). It is \(58.51 \%\) C and \(7.37 \%\) H by mass. Helium effuses through a porous frit \(3.20\) times as fast as the compound does. Determine the empirical and molecular formulas of this compound.

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 .\) L/min at \(1.50\) atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is \(21\) mole percent \(\mathrm{O}_{2}\) and \(79\) mole percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.
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