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Chapter 13: Problem 66

If 2.0 mol of nitrogen gas \(\left(\mathrm{N}_{2}\right)\) are placed in a cubic box, \(25 \mathrm{cm}\) on each side, at 1.6 atm of pressure, what is the rms speed of the nitrogen molecules?

Short Answer

Expert verified
Answer: The root-mean-square speed of the nitrogen molecules is approximately 515.6 m/s.

Step by step solution

01

Write down the given information.

The given information is: - Number of moles of nitrogen gas (\(n\)) = 2.0 moles - Volume of the container (\(V\)) = 25 cm³ on each side = \(25 \times 10^{-3}\) m³ - Pressure of the nitrogen gas (\(P\)) = 1.6 atm = \(1.6 \times 10^5\) Pa (1 atm ≈ \(10^5\) Pa)
02

Use the ideal gas law to find temperature.

Using the ideal gas law \(PV = nRT\), we can find the temperature \(T\): \(T = \frac{PV}{nR}\) where \(R\) is the universal gas constant (\(8.314 \frac{J}{mol\cdot K}\)). Plug in the given values: \(T = \frac{(1.6 \times 10^5 \, Pa)(25 \times 10^{-3} \, m^3)}{(2.0 \, mol)(8.314 \frac{J}{mol \cdot K})}\) Now, calculate the temperature: \(T \approx 241.2 \, K\)
03

Calculate molar mass of nitrogen

Nitrogen has an atomic mass of 14 g/mol. Since each nitrogen molecule is diatomic (\(N_2\)), the molar mass of nitrogen gas is: \(M = 2 \times (14 \frac{g}{mol}) = 28 \frac{g}{mol}\) Convert this molar mass to kilograms per mole: \(M = 28 \times 10^{-3} \, \frac{kg}{mol}\)
04

Use rms speed formula to find the speed of nitrogen molecules.

Now, use the rms speed formula: \(v_{rms} = \sqrt{\frac{3RT}{M}}\) Plug in the values for \(R\), \(T\), and \(M\): \(v_{rms} = \sqrt{\frac{(3) (8.314 \frac{J}{mol \cdot K}) (241.2 \, K)}{28 \times 10^{-3} \, \frac{kg}{mol}}}\) And finally, find the rms speed: \(v_{rms} \approx 515.6 \, \frac{m}{s}\) Hence, the rms speed of the nitrogen molecules in the container is approximately \(515.6 \, \frac{m}{s}\).

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Most popular questions from this chapter

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