Chapter 17: Problem 19
An object must have a speed of at least \(11.2 \mathrm{~km} / \mathrm{s}\) to escape from the Earth's gravitational field. At what temperature will \(v_{\mathrm{rms}}\) for \(\mathrm{H}_{2}\) molecules equal the escape speed? Repeat for \(\mathrm{N}_{2}\) molecules. \(\left(M_{\mathrm{H} 2}=2.0 \mathrm{~kg} / \mathrm{kmol}\right.\) and \(M_{\mathrm{N} 2}=28 \mathrm{~kg} / \mathrm{kmol}\).)
Short Answer
Step by step solution
Understand the Problem
Formula for RMS Velocity
Convert Units
Calculate RMS Velocity for H2
Calculate RMS Velocity for N2
Verify Results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Root-Mean-Square Velocity
- \( v_{\mathrm{rms}} \) is the root-mean-square velocity.
- \( k \) is the Boltzmann constant \( (1.38 \times 10^{-23} \, \text{J/K}) \).
- \( T \) is the temperature in Kelvin.
- \( M \) is the molar mass of the gas, expressed in kilograms per mole.
Escape Velocity
Boltzmann Constant
Kinetic Theory of Gases
- Gas molecules are in constant, random motion and the collisions between them and with the walls of their container are elastic.
- The volume of the individual gas molecules is negligible compared to the total volume of the gas.
- The average kinetic energy of gas molecules is proportional to the absolute temperature of the gas.