Mixing Gases
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
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Exercise:
abcliste abc Using the expression for the partition function Z for an ideal gas but modified to ase that the particles are distinguishable calculate the entropy of the ideal gas using the thermodynamic relation S fracpartialpartial Tk_BTln Z_N V. abc Imagine that we have two identical containers with the same gas of distinguishable particles. Each container has a volume V particle number N and a temperature T . They are separated by a thin wall. Now suppose the wall is removed. What is the change in entropy of the system? abc Now ase that the particles of the ideal gas are indistinguishable. Recalculate the results of parts a and b. Here it is useful to use Stirling’s approximation which says that lnN!approx Nln N-N in the limit of large N with errors on the order of lnN. abcliste
Solution:
abcliste abc We are treating distinguishable particles thus we have to drop the N! factor which is found when treating indistiguishable oes. We thus can write: Z V^Nleftfracpi mk_BTh^right^fracN. Now using the given thermodynamic relation we can derive the entropy of the system: S fracpartialpartial Tk_BTln Z_N V fracpartialpartial Tleftk_BTln left V^Nleftfracpi mk_BTh^right^fracNrightright k_Blnleft V^Nleftfracpi mk_BTh^right^fracNright + frack_BT fracNT^fracN-V^Nleftfracpi mk_BTh^right^fracNV^Nleftfracpi mk_BTh^right^fracN k_Blnleft V^Nleftfracpi mk_BTh^right^fracNright + fracNk_B abc We can use the entropy computed in part a to solve this part of the exercise. In particular Delta S S_textNew - S_textOld SVNT−SVNT. Using Eq. we can rewrite Delta S as: Delta S k_Bln V^N-k_BlnV^N+fracNk_Blnleftfracpi mk_BTh^right - fracNk_B ln leftfracpi mk_BTh^right + fracNk_B - fracNk_B Nk_Bln + ln V - Nk_Bln V Nk_Bln &neq abc Now considering a case with indistinguishable particle we have to reconsider the N!. In particular ln Z_textindist ln Z_textdist-ln N! This implies that we can rewrite the entropy SV N T as S k_BlnleftV^Nleftfracpi mk_BTh^right^fracNright + fracNk_B - k_BlnN!. From this we can also compute the variation in Delta S as in the part b using also the Stirling's approximation Delta S Nk_Bln - k_BlnN!+k_BlnN! &approx Nk_Bln - k_BNlnN+k_BN+k_BNlnN-k_BN abcliste
abcliste abc Using the expression for the partition function Z for an ideal gas but modified to ase that the particles are distinguishable calculate the entropy of the ideal gas using the thermodynamic relation S fracpartialpartial Tk_BTln Z_N V. abc Imagine that we have two identical containers with the same gas of distinguishable particles. Each container has a volume V particle number N and a temperature T . They are separated by a thin wall. Now suppose the wall is removed. What is the change in entropy of the system? abc Now ase that the particles of the ideal gas are indistinguishable. Recalculate the results of parts a and b. Here it is useful to use Stirling’s approximation which says that lnN!approx Nln N-N in the limit of large N with errors on the order of lnN. abcliste
Solution:
abcliste abc We are treating distinguishable particles thus we have to drop the N! factor which is found when treating indistiguishable oes. We thus can write: Z V^Nleftfracpi mk_BTh^right^fracN. Now using the given thermodynamic relation we can derive the entropy of the system: S fracpartialpartial Tk_BTln Z_N V fracpartialpartial Tleftk_BTln left V^Nleftfracpi mk_BTh^right^fracNrightright k_Blnleft V^Nleftfracpi mk_BTh^right^fracNright + frack_BT fracNT^fracN-V^Nleftfracpi mk_BTh^right^fracNV^Nleftfracpi mk_BTh^right^fracN k_Blnleft V^Nleftfracpi mk_BTh^right^fracNright + fracNk_B abc We can use the entropy computed in part a to solve this part of the exercise. In particular Delta S S_textNew - S_textOld SVNT−SVNT. Using Eq. we can rewrite Delta S as: Delta S k_Bln V^N-k_BlnV^N+fracNk_Blnleftfracpi mk_BTh^right - fracNk_B ln leftfracpi mk_BTh^right + fracNk_B - fracNk_B Nk_Bln + ln V - Nk_Bln V Nk_Bln &neq abc Now considering a case with indistinguishable particle we have to reconsider the N!. In particular ln Z_textindist ln Z_textdist-ln N! This implies that we can rewrite the entropy SV N T as S k_BlnleftV^Nleftfracpi mk_BTh^right^fracNright + fracNk_B - k_BlnN!. From this we can also compute the variation in Delta S as in the part b using also the Stirling's approximation Delta S Nk_Bln - k_BlnN!+k_BlnN! &approx Nk_Bln - k_BNlnN+k_BN+k_BNlnN-k_BN abcliste
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Exercise:
abcliste abc Using the expression for the partition function Z for an ideal gas but modified to ase that the particles are distinguishable calculate the entropy of the ideal gas using the thermodynamic relation S fracpartialpartial Tk_BTln Z_N V. abc Imagine that we have two identical containers with the same gas of distinguishable particles. Each container has a volume V particle number N and a temperature T . They are separated by a thin wall. Now suppose the wall is removed. What is the change in entropy of the system? abc Now ase that the particles of the ideal gas are indistinguishable. Recalculate the results of parts a and b. Here it is useful to use Stirling’s approximation which says that lnN!approx Nln N-N in the limit of large N with errors on the order of lnN. abcliste
Solution:
abcliste abc We are treating distinguishable particles thus we have to drop the N! factor which is found when treating indistiguishable oes. We thus can write: Z V^Nleftfracpi mk_BTh^right^fracN. Now using the given thermodynamic relation we can derive the entropy of the system: S fracpartialpartial Tk_BTln Z_N V fracpartialpartial Tleftk_BTln left V^Nleftfracpi mk_BTh^right^fracNrightright k_Blnleft V^Nleftfracpi mk_BTh^right^fracNright + frack_BT fracNT^fracN-V^Nleftfracpi mk_BTh^right^fracNV^Nleftfracpi mk_BTh^right^fracN k_Blnleft V^Nleftfracpi mk_BTh^right^fracNright + fracNk_B abc We can use the entropy computed in part a to solve this part of the exercise. In particular Delta S S_textNew - S_textOld SVNT−SVNT. Using Eq. we can rewrite Delta S as: Delta S k_Bln V^N-k_BlnV^N+fracNk_Blnleftfracpi mk_BTh^right - fracNk_B ln leftfracpi mk_BTh^right + fracNk_B - fracNk_B Nk_Bln + ln V - Nk_Bln V Nk_Bln &neq abc Now considering a case with indistinguishable particle we have to reconsider the N!. In particular ln Z_textindist ln Z_textdist-ln N! This implies that we can rewrite the entropy SV N T as S k_BlnleftV^Nleftfracpi mk_BTh^right^fracNright + fracNk_B - k_BlnN!. From this we can also compute the variation in Delta S as in the part b using also the Stirling's approximation Delta S Nk_Bln - k_BlnN!+k_BlnN! &approx Nk_Bln - k_BNlnN+k_BN+k_BNlnN-k_BN abcliste
abcliste abc Using the expression for the partition function Z for an ideal gas but modified to ase that the particles are distinguishable calculate the entropy of the ideal gas using the thermodynamic relation S fracpartialpartial Tk_BTln Z_N V. abc Imagine that we have two identical containers with the same gas of distinguishable particles. Each container has a volume V particle number N and a temperature T . They are separated by a thin wall. Now suppose the wall is removed. What is the change in entropy of the system? abc Now ase that the particles of the ideal gas are indistinguishable. Recalculate the results of parts a and b. Here it is useful to use Stirling’s approximation which says that lnN!approx Nln N-N in the limit of large N with errors on the order of lnN. abcliste
Solution:
abcliste abc We are treating distinguishable particles thus we have to drop the N! factor which is found when treating indistiguishable oes. We thus can write: Z V^Nleftfracpi mk_BTh^right^fracN. Now using the given thermodynamic relation we can derive the entropy of the system: S fracpartialpartial Tk_BTln Z_N V fracpartialpartial Tleftk_BTln left V^Nleftfracpi mk_BTh^right^fracNrightright k_Blnleft V^Nleftfracpi mk_BTh^right^fracNright + frack_BT fracNT^fracN-V^Nleftfracpi mk_BTh^right^fracNV^Nleftfracpi mk_BTh^right^fracN k_Blnleft V^Nleftfracpi mk_BTh^right^fracNright + fracNk_B abc We can use the entropy computed in part a to solve this part of the exercise. In particular Delta S S_textNew - S_textOld SVNT−SVNT. Using Eq. we can rewrite Delta S as: Delta S k_Bln V^N-k_BlnV^N+fracNk_Blnleftfracpi mk_BTh^right - fracNk_B ln leftfracpi mk_BTh^right + fracNk_B - fracNk_B Nk_Bln + ln V - Nk_Bln V Nk_Bln &neq abc Now considering a case with indistinguishable particle we have to reconsider the N!. In particular ln Z_textindist ln Z_textdist-ln N! This implies that we can rewrite the entropy SV N T as S k_BlnleftV^Nleftfracpi mk_BTh^right^fracNright + fracNk_B - k_BlnN!. From this we can also compute the variation in Delta S as in the part b using also the Stirling's approximation Delta S Nk_Bln - k_BlnN!+k_BlnN! &approx Nk_Bln - k_BNlnN+k_BN+k_BNlnN-k_BN abcliste
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