Wednesday 12 February 2014

One mole of Argon gas is confined to a `1.0 L` container at a pressure of `10 atm` . What is the rms speed of the argon atoms?


We can express the rms speed of Argon atoms by combining the ideal gas law with the equation for the root mean square velocity of an atom.


The ideal gas law is:


`PV=nRT`


The rms velocity of a molecule with molar mass `M` is:


`v_(rms)=sqrt((3RT)/M)`


From the ideal gas law we have:


`RT=(PV)/n`


Substitute this into the `v_(rms)` relation.


`v_(rms)=sqrt((3PV)/(nM))`


Substitute numerical values and evaluate `v_(rms)` for Argon atoms.


`v_(rms)=sqrt((3(10 atm)(101.325 (kPa)/(atm))(1.0*10^-3 m^3))/((1 mol)(39.948*10^-3 (kg)/(mol))))`


...



We can express the rms speed of Argon atoms by combining the ideal gas law with the equation for the root mean square velocity of an atom.


The ideal gas law is:


`PV=nRT`


The rms velocity of a molecule with molar mass `M` is:


`v_(rms)=sqrt((3RT)/M)`


From the ideal gas law we have:


`RT=(PV)/n`


Substitute this into the `v_(rms)` relation.


`v_(rms)=sqrt((3PV)/(nM))`


Substitute numerical values and evaluate `v_(rms)` for Argon atoms.


`v_(rms)=sqrt((3(10 atm)(101.325 (kPa)/(atm))(1.0*10^-3 m^3))/((1 mol)(39.948*10^-3 (kg)/(mol))))`


`v_(rms)=0.28 (km)/s`


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