Problem 108
Question
The average energy and the rms speed of molecules in a sample of oxygen gas at \(400 \mathrm{~K}\) are \(7.21 \times 10^{-21} \mathrm{~J}\) and \(524 \mathrm{~ms}^{-1}\) respectively. The corresponding values at \(800 \mathrm{~K}\) are nearly (a) \(14.42 \times 10^{-21} 3,1048 \mathrm{~ms}^{-1}\) (b) \(10.18 \times 10^{-21} J, 741 \mathrm{~ms}^{-1}\) (c) \(7.21 \times 10^{-21} \mathrm{~J}, 1048 \mathrm{~ms}^{-1}\) (d) \(14.42 \times 10^{-21} \mathrm{~J}, 741 \mathrm{~ms}^{-1}\)
Step-by-Step Solution
Verified Answer
The correct option is (d) \(14.42 \times 10^{-21} \mathrm{~J}, 741 \mathrm{~ms}^{-1}\).
1Step 1: Calculate New Average Energy
The average energy of a molecule in a gas is proportional to the absolute temperature (T). If the temperature doubles from 400 K to 800 K, the average energy should also double. \ The given average energy at 400 K is \(7.21 \times 10^{-21} \mathrm{~J}\). Therefore, at 800 K, the new average energy will be: \ \[2 \times 7.21 \times 10^{-21} = 14.42 \times 10^{-21} \mathrm{~J}\]
2Step 2: Calculate New RMS Speed
The root mean square (rms) speed of gas molecules is given by the formula: \ \[ v_{rms} = \sqrt{\frac{3kT}{m}} \] According to the formula, \(v_{rms}\) is proportional to the square root of the temperature \(T\). When the temperature doubles, \(v_{rms}\) will change by a factor of \(\sqrt{2}\). \ The given rms speed at 400 K is \(524 \mathrm{~ms}^{-1}\). Therefore, at 800 K, the new rms speed will be: \ \[v_{rms} = 524 \times \sqrt{2} \approx 741 \mathrm{~ms}^{-1}\]
3Step 3: Compare Against Options
The calculated average energy at 800 K is \(14.42 \times 10^{-21} \mathrm{~J}\) and the rms speed is \(741 \mathrm{~ms}^{-1}\). Comparing these results with the provided options: \ (a) \(14.42 \times 10^{-21} 3,1048 \mathrm{~ms}^{-1}\) - Incorrect rms speed. \ (b) \(10.18 \times 10^{-21} J, 741 \mathrm{~ms}^{-1}\) - Incorrect average energy. \ (c) \(7.21 \times 10^{-21} \mathrm{~J}, 1048 \mathrm{~ms}^{-1}\) - Incorrect in both. \ (d) \(14.42 \times 10^{-21} \mathrm{~J}, 741 \mathrm{~ms}^{-1}\) - Correct. \
Key Concepts
Average EnergyRMS SpeedTemperature Dependence
Average Energy
In the Kinetic Theory of Gases, the average energy of a gas molecule is directly related to the temperature of the gas. As a gas heats up, the energy of each molecule increases proportionately. The formula connecting temperature and average energy is: \[ E_{avg} = \frac{3}{2}kT \] where \( E_{avg} \) represents the average kinetic energy of the molecules, \( k \) is the Boltzmann constant, and \( T \) is the absolute temperature.
- When the temperature doubles, the average energy also doubles.
- This relationship shows how energy per molecule increases as heat is applied to the gas.
RMS Speed
The Root Mean Square (RMS) speed is another important aspect of the kinetic theory concerning gases. It describes the speed of gas molecules, determining the broad motion and energetics of the gas. RMS speed can be calculated using the formula: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \] where \( v_{rms} \) stands for the root mean square speed, \( k \) is the Boltzmann constant, \( T \) is the absolute temperature, and \( m \) is the mass of a gas molecule.
- The RMS speed is directly tied to the temperature, influenced by the square root of the temperature.
- When the temperature doubles, the RMS speed increases by a factor of \(\sqrt{2}\).
Temperature Dependence
Understanding temperature dependence is crucial in the Kinetic Theory of Gases. It articulates how various properties of gases, such as average energy and RMS speed, evolve with changes in temperature. Let's examine why temperature plays such a vital role:
- Temperature reflects the internal energy, so an increase means the molecules move faster and have more kinetic energy.
- Proportional changes in temperature cause proportional changes in average energy, whereas RMS speed is adjusted by the square root of temperature changes.
- The average kinetic energy of molecules directly doubles.
- The RMS speed increases by the factor \(\sqrt{2}\).
Other exercises in this chapter
Problem 106
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