- #1
Faiq
- 348
- 16
Homework Statement
Why is the formula ##p = \frac{1}{3}\rho<c^2>## used to calculate the mean square speed at 273K?
Why 273K?
Gas Laws -- why calculate the mean square speed at 273K?
In summary, the equation used to calculate the mean square speed at 273K is ##p = \frac{1}{3}\rho<c^2>##. This equation is applicable to any temperature given the given pressure and density.
Why is the formula ##p = \frac{1}{3}\rho<c^2>## used to calculate the mean square speed at 273K?
Why 273K?
- #2
BvU
Science Advisor
Homework Helper
- 16,095
- 4,863
It works at any temperature. Could you give a bit more context ? Is this a section in your textbook you have a question about or is it in an exercise ?
- #3
Faiq
- 348
- 16
The density of a gas at a temperature of ##273~K## and a pressure of ##1.02*10^5~ Pa## is ##0.9kgm^{-3}##. It may be asssumed to be an ideal gas.
Calculate rms speed at 273K and 546K.
For 273, we were supposed to use the given equation and for second we were supposed to used the ratio of mean square speed and temperature.
My thinking is the equation only applies for those temperature for which we are using the pressure values.
Calculate rms speed at 273K and 546K.
For 273, we were supposed to use the given equation and for second we were supposed to used the ratio of mean square speed and temperature.
My thinking is the equation only applies for those temperature for which we are using the pressure values.
- #4
Borek
Mentor
- 29,042
- 4,413
You are given pressure and density at 273 K and asked to calculate rms at exactly this temperature, I don't see where is the problem?
- #5
Faiq
- 348
- 16
The confusion I had was why is that formula only applicable for the 273K. That's to say why I can't just put that value down for 546K? Later I realized, the pressure inserted in those equations are the indicator of what temperature should the RMS value be for.Borek said:
- #6
BvU
Science Advisor
Homework Helper
- 16,095
- 4,863
We're still missing context. Please be a bit more complete in formulating problem statements. This way we spend way too much time dragging out what is troubling you.
Accidentally ? Or did you erase them ? That's a no-no in PF !
And anyway, 1. should probably have been
Did you do anything at all with the information that the gas can be considered an ideal gas ?
Where did the
?)
Missing:Faiq said:
Homework Equations
The Attempt at a Solution
Accidentally ? Or did you erase them ? That's a no-no in PF !
And anyway, 1. should probably have been
Any word or sentence missing from this full problem statement ? If no, then a list of all variables is good to have at hand. For you too.Faiq said:
What given equation ?Faiq said:
Did you do anything at all with the information that the gas can be considered an ideal gas ?
Where did the
come from ? From the textbook or from the problem statement ? (or perhaps from the solution manualFaiq said:
?)- #7
Faiq
- 348
- 16
1. No words or sentences are missing.
2. The given equation is ##p = \frac{1}{3}\rho<c^2>##
3. By telling us that it could be considered an ideal gas, permission was given to use the ideal gas law and the provided equation.
4. The formula was supposed to be derived by using the equation ##pV = \frac{1}{3}Nm<c^2>## and ##\rho = \frac{Nm}{V}##
Btw my confusion has been sorted out. Thanks for the help though
2. The given equation is ##p = \frac{1}{3}\rho<c^2>##
3. By telling us that it could be considered an ideal gas, permission was given to use the ideal gas law and the provided equation.
4. The formula was supposed to be derived by using the equation ##pV = \frac{1}{3}Nm<c^2>## and ##\rho = \frac{Nm}{V}##
Btw my confusion has been sorted out. Thanks for the help though
FAQ: Gas Laws -- why calculate the mean square speed at 273K?
1. Why is it important to calculate the mean square speed at 273K in gas laws?
The mean square speed at 273K is an important parameter in gas laws because it is directly related to the temperature of the gas. It helps us understand the kinetic energy of gas particles, which in turn affects the pressure, volume, and temperature of the gas.
2. What does the mean square speed at 273K represent?
The mean square speed at 273K represents the average speed of gas particles at a temperature of 273 Kelvin. It is calculated by taking the square root of the average of the squared speeds of individual gas particles.
3. How is the mean square speed at 273K related to the root mean square speed?
The mean square speed at 273K is equal to three times the root mean square speed. This relationship allows us to easily calculate the mean square speed if we know the root mean square speed or vice versa.
4. How does the mean square speed at 273K change with temperature?
The mean square speed at 273K increases as the temperature of the gas increases. This is because as temperature increases, the kinetic energy of gas particles also increases, leading to higher speeds and a larger mean square speed.
5. What is the significance of using 273K in calculating the mean square speed?
273K is the standard temperature at which gas laws are typically studied. By using this temperature, we can compare the mean square speeds of different gases and determine their relative speeds and kinetic energies. It also allows for easier conversion between different temperature units, such as Celsius and Kelvin.
Similar threads
Chemistry
Ideal Gas Law and Partial Pressures