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Chapter 19: Problem 5

Which of the following gases has the highest rootmean-square speed? a) nitrogen at \(1 \mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) b) argon at \(1 \mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) c) argon at \(2 \mathrm{~atm}\) and \(30^{\circ} \mathrm{C}\) d) oxygen at 2 atm and \(30^{\circ} \mathrm{C}\) e) nitrogen at \(2 \mathrm{~atm}\) and \(15^{\circ} \mathrm{C}\)

Short Answer

Expert verified
a) Nitrogen at 1 atm and 30掳C b) Argon at 1 atm and 30掳C c) Argon at 2 atm and 30掳C d) Oxygen at 2 atm and 30掳C e) Nitrogen at 2 atm and 15掳C Answer: (a) Nitrogen at 1 atm and 30掳C

Step by step solution

01

Convert temperatures to Kelvin

To convert temperatures from Celsius to Kelvin, we add 273.15. \(T_1 = 30 + 273.15 = 303.15 \mathrm{~K}\) \(T_2 = 15 + 273.15 = 288.15 \mathrm{~K}\)
02

Find the molar masses of the gases

Molar Mass of Nitrogen (N\(_2\)) = \(28.02 \mathrm{~g/mol}\) Molar Mass of Argon (Ar) = \(39.95 \mathrm{~g/mol}\) Molar Mass of Oxygen (O\(_2\)) = \(32.00 \mathrm{~g/mol}\)
03

Calculate the root-mean-square speeds

We use the ideal gas constant in J/(mol K): \(R = 8.314 \mathrm{~J/(mol\cdot K)}\) a) Nitrogen at 1 atm and 303.15 K: \(v_\mathrm{rms,a} = \sqrt{\frac{3(8.314)(303.15)}{28.02}} = 515.3 \mathrm{~m/s}\) b) Argon at 1 atm and 303.15 K: \(v_\mathrm{rms,b} = \sqrt{\frac{3(8.314)(303.15)}{39.95}} = 431.6 \mathrm{~m/s}\) c) Argon at 2 atm and 303.15 K: \(v_\mathrm{rms,c} = \sqrt{\frac{3(8.314)(303.15)}{39.95}} = 431.6 \mathrm{~m/s}\) d) Oxygen at 2 atm and 303.15 K: \(v_\mathrm{rms,d} = \sqrt{\frac{3(8.314)(303.15)}{32.00}} = 482.4 \mathrm{~m/s}\) e) Nitrogen at 2 atm and 288.15 K: \(v_\mathrm{rms,e} = \sqrt{\frac{3(8.314)(288.15)}{28.02}} = 501.5 \mathrm{~m/s}\)
04

Compare the root-mean-square speeds

Comparing the root-mean-square speeds, we find that: \(v_\mathrm{rms,a} > v_\mathrm{rms,b}\), \(v_\mathrm{rms,c}\), \(v_\mathrm{rms,d}\), and \(v_\mathrm{rms,e}\) Thus, the gas with the highest root-mean-square speed is option (a) nitrogen at 1 atm and 30掳C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Ideal Gas Constant
The ideal gas constant is an essential concept when dealing with gases. It is symbolized by \( R \) and has a value of \( 8.314 \text{ J/(mol}\cdot\text{K)} \). The ideal gas constant links the temperature, pressure, and volume of an ideal gas. It appears in the ideal gas law equation: \( PV = nRT \). Here, \( P \) represents pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature measured in Kelvin.
  • The unit J/(mol K) reflects the linkage with energy, moles, and temperature.
  • In root-mean-square speed calculations, \( R \) helps determine the speed of gas particles based on the temperature and molar mass.
  • Remember that \( R \) remains constant for an ideal gas under various conditions, simplifying many calculations.
Understanding \( R \) allows one to connect various gas properties and make relevant predictions using calculations like the root-mean-square speed, which reflects the average speed of gas molecules at a given temperature.
Importance of Temperature Conversion
When dealing with gas calculations, it is crucial to convert temperatures from Celsius to Kelvin. The Kelvin scale is an absolute temperature scale and is used in calculations involving the ideal gas constant. Converting Celsius to Kelvin is simple: add 273.15 to the Celsius temperature.
  • This conversion ensures consistency, as the ideal gas law and other related formulas require temperatures in Kelvin.
  • Kelvin is preferred in scientific calculations because it starts at absolute zero, simplifying entropy and thermodynamics equations.
  • In our exercise, the temperatures 30掳C and 15掳C were converted to 303.15 K and 288.15 K, respectively, to facilitate further calculations.
Ensuring correct temperature conversion eliminates errors in gas law applications and influences calculations like root-mean-square speed, which rely on accurate temperature inputs.
Calculating Molar Mass for Gas Studies
Molar mass plays a pivotal role in calculating physical properties of gases, such as root-mean-square speed. It represents the mass of one mole of a substance, expressed in grams per mole (g/mol). To find the molar mass:
  • For diatomic molecules like nitrogen (N\(_2\)) and oxygen (O\(_2\)), add the atomic masses of the atoms.
  • For instance, nitrogen's molar mass is \( 28.02 \text{ g/mol} \) and oxygen's molar mass is \( 32.00 \text{ g/mol} \).
  • Argon, being a monoatomic molecule, has a molar mass of \( 39.95 \text{ g/mol} \).
Calculating molar mass correctly is key in determining a gas's root-mean-square speed. This speed decreases with larger molar masses, reflecting the inverse relationship between mass and velocity. Thus, by knowing the molar mass, we can predict how fast molecules are moving based on the root-mean-square formula.

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