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Chapter 22: Problem 16

Calculate the root-mean-square speed of ammonia \(\left(\mathrm{NH}_{3}\right)\) molecules at \(56.0^{\circ} \mathrm{C}\). An atom of nitrogen has a mass of \(2.33 \times 10^{-26} \mathrm{~kg}\) and an atom of hydrogen has a mass of \(1.67 \times 10^{-27} \mathrm{~kg}\)

Short Answer

Expert verified
The root-mean-square speed of the ammonia molecule at \(56.0^{\circ}\) is approximately \(665 m/s\)

Step by step solution

01

Convert Temperature from Celsius to Kelvin

To convert temperature from Celsius to Kelvin, the following formula is used: \[T(K) = T(^{\circ}C) + 273.15\]. Substitute the given temperature (56.0°C) to get \[T = 56.0 + 273.15 = 329.15 K\]
02

Calculate Mass of One Ammonia Molecule

One molecule of ammonia (NH3) has one Nitrogen atom and three Hydrogen atoms. The total mass is hence the sum of the mass of each atom. The given masses are Nitrogen: \(2.33 \times 10^{-26}\) kg and Hydrogen: \(1.67 \times 10^{-27}\) kg. So total mass is \[m = 2.33 \times 10^{-26} kg + 3 \times 1.67 \times 10^{-27} kg = 2.33 \times 10^{-26} kg + 5.01 \times 10^{-27} kg = 2.83 \times 10^{-26} kg\]
03

Calculate Root-Mean-Square Speed

Now we have the temperature in Kelvin, the Boltzmann constant, and the mass of one ammonia molecule. We can substitute these into the formula \[v_{rms}=\sqrt{\frac{3kT}{m}}\]. Using the Boltzmann constant \(k = 1.38 \times 10^{-23} m^{2} kg s^{-2} K^{-1}\) , we get \[v_{rms}=\sqrt{\frac{3 \times 1.38 \times 10^{-23} m^{2} kg s^{-2} K^{-1} \times 329.15 K}{2.83 \times 10^{-26} kg}} = 664.97 m/s\]

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Most popular questions from this chapter

The speeds of a group of ten molecules are \(2.0,3.0,4.0, \ldots\), \(11 \mathrm{~km} / \mathrm{s}\). ( \(a\) ) Find the average speed of the group. (b) Calculate the root-mean-square speed of the group.

At \(44.0^{\circ} \mathrm{C}\) and \(1.23 \times 10^{-2}\) atm the density of a gas is \(1.32 \times 10^{-5} \mathrm{~g} / \mathrm{cm}^{3} .(a)\) Find \(v_{\mathrm{rms}}\) for the gas molecules. (b) Using the ideal gas law, find the number of moles per unit volume (molar density) of the gas. ( \(c\) ) By combining the results of (a) and \((b)\), find the molar mass of the gas and identify it.

Estimate the van der Waals constant \(b\) for \(\mathrm{H}_{2} \mathrm{O}\) knowing that one kilogram of water has a volume of \(0.001 \mathrm{~m}^{3}\). The molar mass of water is \(18 \mathrm{~g} / \mathrm{mol}\).

You are given the following group of particles \(\left(N_{n}\right.\) represents the number of particles that have a speed \(v_{n}\) ): $$\begin{array}{lc}N_{n} & v_{n}(\mathrm{~km} / \mathrm{s}) \\\\\hline 2 & 1.0 \\\4 & 2.0 \\\6 & 3.0 \\\8 & 4.0 \\\2 & 5.0\end{array}$$ (a) Compute the average speed \(v_{\mathrm{av}} .(b)\) Compute the rootmean- square speed \(v_{\text {rms. }}\) (c) Among the five speeds shown, which is the most probable speed \(v_{\mathrm{p}}\) for the entire group?

(a) Ten particles are moving with the following speeds: four at \(200 \mathrm{~m} / \mathrm{s}\), two at \(500 \mathrm{~m} / \mathrm{s}\), and four at \(600 \mathrm{~m} / \mathrm{s}\). Calculate the average and root-mean-square speeds. Is \(v_{\mathrm{rms}}>v_{\mathrm{av}} ?(b)\) Make up your own speed distribution for the ten particles and show that \(v_{\mathrm{rms}} \geq v_{\mathrm{av}}\) for your distribution. ( \(c\) ) Under what condition (if any) does \(v_{\mathrm{rms}}=v_{\mathrm{av}} ?\)
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