Faiq
- 347
- 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?
The discussion revolves around the application of the formula ##p = \frac{1}{3}\rho
Participants express differing views on the applicability of the formula at various temperatures, with some asserting it can be used universally while others believe it is limited to specific conditions. The discussion remains unresolved regarding the implications of using the formula at temperatures other than 273K.
Some participants highlight the importance of understanding the context of the problem, including the assumptions made about the gas being ideal and the specific conditions under which the formula is applied. There are references to missing details in the problem statement that could clarify the situation further.
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: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?
Missing:Faiq said:Homework Statement
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: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.
What given equation ?Faiq said: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.
come from ? From the textbook or from the problem statement ? (or perhaps from the solution manualFaiq said:formula ##p = \frac{1}{3}\rho<c^2>##