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There are two conceptual questions from last year's exam that I don't know the answer to.
1) From the point of vue of statistical mechanics, why can't we completely neglect the interaction btw particles, even in an ideal gas?
My best shot is "Though we can neglect the potential energy btw the particles, we still have to account for the collisions btw them." But I don't recall of a place in what we have covered so far about statistical mechanics, where we did account for the collisions btw the particles of a system.
2) Why is the specific heat of a polyatomic gas greater then that of a monoatomic one?
This information is contained within the text, but upon further inspection, I don't see how the explanation given explains the phenomenon. The explanation given in the text is that the polyatomic molecules have rotational degrees of freedom additionally to their translation degree of freedom, so a part of the energy goes into rotational energy, which considerably augment the number of accessible states.
But the specific heat is given by C=dQ/dT. Alright, so say a quantity dQ of heat is transferred to the polyatomic gaz. Then its energy goes up by dQ. The temperature of the gaz is defined by
kT\equiv \frac{1}{\frac{\partial \ln(\Omega)}{\partial E}}
(where \Omega is the number of accesible states). We also know that for an ideal gaz, T is function of E only, so
kdT = \frac{dT}{dE}dE = \frac{d}{dE}\left( \frac{1}{\frac{\partial \ln(\Omega)}{\partial E}}\right) dE = -\frac{1}{\left( \frac{\partial \ln(\Omega)}{\partial E} \right)^2} \frac{\partial^2 \ln(\Omega)}{\partial E^2}dE
Granted, \frac{\partial \ln(\Omega)}{\partial E} is greater for a polyatomic gaz than for a monoatomic one, but I know nothing about the second derivative of \ln(\Omega), other than it must be negative for dT to be positive as it should. So, is there something I don't see?Btw - I would appreciate an answer before 12 hours from now, because the exam is in 12h30 and I live 15 minutes away. :-)
1) From the point of vue of statistical mechanics, why can't we completely neglect the interaction btw particles, even in an ideal gas?
My best shot is "Though we can neglect the potential energy btw the particles, we still have to account for the collisions btw them." But I don't recall of a place in what we have covered so far about statistical mechanics, where we did account for the collisions btw the particles of a system.
2) Why is the specific heat of a polyatomic gas greater then that of a monoatomic one?
This information is contained within the text, but upon further inspection, I don't see how the explanation given explains the phenomenon. The explanation given in the text is that the polyatomic molecules have rotational degrees of freedom additionally to their translation degree of freedom, so a part of the energy goes into rotational energy, which considerably augment the number of accessible states.
But the specific heat is given by C=dQ/dT. Alright, so say a quantity dQ of heat is transferred to the polyatomic gaz. Then its energy goes up by dQ. The temperature of the gaz is defined by
kT\equiv \frac{1}{\frac{\partial \ln(\Omega)}{\partial E}}
(where \Omega is the number of accesible states). We also know that for an ideal gaz, T is function of E only, so
kdT = \frac{dT}{dE}dE = \frac{d}{dE}\left( \frac{1}{\frac{\partial \ln(\Omega)}{\partial E}}\right) dE = -\frac{1}{\left( \frac{\partial \ln(\Omega)}{\partial E} \right)^2} \frac{\partial^2 \ln(\Omega)}{\partial E^2}dE
Granted, \frac{\partial \ln(\Omega)}{\partial E} is greater for a polyatomic gaz than for a monoatomic one, but I know nothing about the second derivative of \ln(\Omega), other than it must be negative for dT to be positive as it should. So, is there something I don't see?Btw - I would appreciate an answer before 12 hours from now, because the exam is in 12h30 and I live 15 minutes away. :-)
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