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Chapter 6: Problem 80

Calculate the total kinetic energy, in joules, of \(155 \mathrm{g} \mathrm{N}_{2}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and 1.00 atm. \([\text {Hint}:\) First calculate the average kinetic energy, \(\bar{e}_{k}\).

Short Answer

Expert verified
Therefore, the total kinetic energy of \(155 g\) of Nitrogen gas at \(25^{\circ}C\) and \(1.00 atm\) is \(2.05 \times 10^4 J\).

Step by step solution

01

Find Temperature in Kelvin

Convert the given temperature from Celsius to Kelvin using the relation: \(T(K) = T(^\circ C) + 273.15\). So, for \(T = 25^\circ C\), the temperature in Kelvin is \(T = 25 + 273.15 = 298.15 K\).
02

Calculate Average Kinetic Energy

Use the formula for the average kinetic energy, which is \(\bar{e}_{k} = \frac{3}{2}kT\). Substituting Boltzmann's constant \(k = 1.38 \times 10^{-23} J/K\) and calculated temperature \(T = 298.15 K\), you find \(\bar{e}_{k} = \frac{3}{2} \times 1.38 \times 10^{-23} J/K \times 298.15 K = 6.17 \times 10^{-21} J\).
03

Find the Number of Molecules

To find the number of nitrogen molecules, first convert the given mass from grams to moles using the molar mass of nitrogen gas (\(28.02 g/mol\)). Then, multiply by Avogadro's number to get the total number of molecules. So, for \(155 g\) of nitrogen gas, the total number of molecules is \((155 g / 28.02 g/mol) \times 6.022 \times 10^{23} mol^{-1} = 3.33 \times 10^{24}\).
04

Find the Total Kinetic Energy

Now multiply the average kinetic energy per molecule by the total number of molecules to find the total kinetic energy, which is \(3.33 \times 10^{24} \times 6.17 \times 10^{-21} J = 2.05 \times 10^4 J\).

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Most popular questions from this chapter

At \(0^{\circ} \mathrm{C}\) and 0.500 atm, 4.48 L of gaseous \(\mathrm{NH}_{3}\) (a) contains \(6.02 \times 10^{22}\) molecules; (b) has a mass of \(17.0 \mathrm{g} ;(\mathrm{c})\) contains \(0.200 \mathrm{mol} \mathrm{NH}_{3} ;\) (d) has a mass of \(3.40 \mathrm{g}\).

The density of phosphorus vapor is \(2.64 \mathrm{g} / \mathrm{L}\) at \(310^{\circ} \mathrm{C}\) and \(775 \mathrm{mm}\) Hg. What is the molecular formula of the phosphorus under these conditions?

The gas with the greatest density at STP is (a) \(\mathrm{N}_{2} \mathrm{O}\) (b) \(\mathrm{Kr} ;\) (c) \(\mathrm{SO}_{3} ;\) (d) \(\mathrm{Cl}_{2}\).

A nitrogen molecule ( \(\mathrm{N}_{2}\) ) having the average kinetic energy at \(300 \mathrm{K}\) is released from Earth's surface to travel upward. If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give your answer in kilometers. [Hint: When the molecule comes to rest, the potential energy of the molecule will be \(m g h\) where \(m\) is the molecular mass in kilograms, \(g=9.81 \mathrm{m} \mathrm{s}^{-2}\) is the acceleration due to gravity, and \(h\) is the height, in meters, above Earth's surface.]

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(23^{\circ} \mathrm{C}\) and a barometric pressure of 751 Torr. The vapor pressure of water at \(23^{\circ} \mathrm{C}\) is \(21 \mathrm{mmHg}\). The partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in the sample collected is (a) \(21 \mathrm{mmHg}_{i}\) (b) 751 Torr; \((\mathrm{c}) 0.96 \mathrm{atm} ;\) (d) \(1.02 \mathrm{atm}\).
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