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Chapter 19: Problem 66

Calculate the root-mean-square speed of air molecules at room temperature \(\left(22.0^{\circ} \mathrm{C}\right)\) from the kinetic theory of an ideal gas.

Short Answer

Expert verified
Answer: The root-mean-square speed of air molecules at room temperature (22.0°C) is approximately 492 m/s.

Step by step solution

01

Write down the given information

Temperature (T) is given as \(22.0^{\circ} \mathrm{C}\). In order to use this in the rms speed formula, we need to convert it to Kelvin, by adding 273.15 to the Celsius value: \(T = 22.0 + 273.15 = 295.15 \, K\). Also, the molar mass (\(M\)) of air needs to be determined. Dry air is composed of approximately 78% nitrogen (\(N_2\)), 21% oxygen (\(O_2\)), and 1% argon (\(Ar\)). We will assume an average molar mass for air based on these percentages, with \(M_{N_2} = 28.02 \, g/mol\), \(M_{O_2} = 32.00 \, g/mol\), and \(M_{Ar} = 39.95 \, g/mol\).
02

Calculate the average molar mass of air

Based on the percentages of nitrogen, oxygen, and argon in air, we can calculate the average molar mass of air using a weighted average: \(M_{air} = 0.78 \times M_{N_2} + 0.21 \times M_{O_2} + 0.01 \times M_{Ar}\) \(M_{air} = 0.78 \times 28.02 + 0.21 \times 32.00 + 0.01 \times 39.95\) \(M_{air} \approx 28.97 \, g/mol\)
03

Use the molar mass to calculate the mass of one air molecule

The formula for root-mean-square speed requires the mass of one molecule (\(m\)). We can obtain this value by dividing the molar mass by the Avogadro constant (\(N_A\)): \(m = \frac{M_{air}}{N_A}\) \(m = \frac{28.97 \, g/mol}{6.022 \times 10^{23} \, molecules/mol}\) \(m \approx 4.81 \times 10^{-26} \, kg\)
04

Write the rms speed formula and identify known values

The root-mean-square (rms) speed, denoted as \(v_{rms}\), is calculated by the following formula: \(v_{rms} = \sqrt{\frac{3kT}{m}}\) Here, \(k\) is the Boltzmann constant (1.38 × 10^{-23} J/K) and \(T\) is the temperature in Kelvin (which we have already converted, and \(m\) is the mass of one molecule, already calculated in previous steps.
05

Calculate the rms speed

By substituting the known values into the equation, we can calculate the rms speed of air molecules at room temperature: \(v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \, J/K \times 295.15 \, K}{4.81 \times 10^{-26} \, kg}}\) \(v_{rms} \approx 492 \, m/s\) The root-mean-square speed of air molecules at room temperature (22.0°C) is approximately 492 m/s.

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Most popular questions from this chapter

Molar specific heat at constant pressure, \(C_{p}\), is larger than molar specific heat at constant volume, \(C_{V}\), for a) a monoatomic ideal gas. b) a diatomic atomic gas. c) all of the above. d) none of the above.

Suppose 5.0 moles of an ideal monatomic gas expand at a constant temperature of \(22^{\circ} \mathrm{C}\) from an initial volume of \(2.0 \mathrm{~m}^{3}\) to \(8.0 \mathrm{~m}^{3}\) a) How much work is done by the gas? b) What is the final pressure of the gas?

Suppose \(15.0 \mathrm{~L}\) of an ideal monatomic gas at a pressure of \(1.50 \cdot 10^{5} \mathrm{kPa}\) is expanded adiabatically (no heat transfer) until the volume is doubled. a) What is the pressure of the gas at the new volume? b) If the initial temperature of the gas was \(300 . \mathrm{K},\) what is its final temperature after the expansion?

An ideal gas has a density of \(0.0899 \mathrm{~g} / \mathrm{L}\) at \(20.00^{\circ} \mathrm{C}\) and \(101.325 \mathrm{kPa}\). Identify the gas.

At room temperature, identical gas cylinders contain 10 moles of nitrogen gas and argon gas, respectively. Determine the ratio of energies stored in the two systems. Assume ideal gas behavior.
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