Asked by niharika
at what temperature will the root mean square velocity of sulphur dioxide be same as that of methane at 27c?
Answers
Answered by
bobpursley
SO2 vs Methane
Methane: symettrical molecule, one degree of freedom (translational)
SO2: linear molecule, two degrees of freedom (translational, vibration).
One of the fundamental postulates in statistical thermodynamics, is that energy <i>almost</i> divides equally in each degree of freedom.
So 1/2 of the energy in SO2 goes to translational energy, and
KE then = 1/2 energy available at a temp, or
energy available= 2*KE
(rmsVelocity)^2 proportional to KE, so
rmsVelocity proportional to sqrt 1/2*KE available
rmsVelocity proportional to sqrt (.5* total energy) or
rmsVelocity proportional to .707 sqrttotal energy
finally, total energy proportional to abs temp.
so rmsVelocity to be the same is .707*sqrt(temp)/sqrt(300)
or temp=2*300K or 600K or 600-273 C
check my thinking. I am a bit surprised at this question, so I assume you are studying statistical thermodynamics.
check my work carefully.
Methane: symettrical molecule, one degree of freedom (translational)
SO2: linear molecule, two degrees of freedom (translational, vibration).
One of the fundamental postulates in statistical thermodynamics, is that energy <i>almost</i> divides equally in each degree of freedom.
So 1/2 of the energy in SO2 goes to translational energy, and
KE then = 1/2 energy available at a temp, or
energy available= 2*KE
(rmsVelocity)^2 proportional to KE, so
rmsVelocity proportional to sqrt 1/2*KE available
rmsVelocity proportional to sqrt (.5* total energy) or
rmsVelocity proportional to .707 sqrttotal energy
finally, total energy proportional to abs temp.
so rmsVelocity to be the same is .707*sqrt(temp)/sqrt(300)
or temp=2*300K or 600K or 600-273 C
check my thinking. I am a bit surprised at this question, so I assume you are studying statistical thermodynamics.
check my work carefully.