{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# sol3 - CHEM 444 HW#3 Answer key by Jiahao Chen due Grading...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHEM 444 HW#3 Answer key by Jiahao Chen due September 24, 2004 Grading policy: One point is given for each numbered equation shown in the solution, or for written state- ments and/or equations to the same effect. The total score for this problem set is 48 points plus a maximum of 3 bonus points. 1. (17-32) Because the molecules in an ideal gas are independent, the partition function of a mixture of monatomic ideal gases is of the form Q ( N 1 ,N 2 ,V,T ) = [ q 1 ( V,T )] N 1 N 1 ! [ q 2 ( V,T )] N 2 N 2 ! (1) where q j ( V,T ) = 2 mk B T h 2 3 2 V (2) Show that h E i = 3 2 ( N 1 + N 2 ) k B T (3) and that PV = ( N 1 + N 2 ) k B T (4) for a mixture of monatomic ideal gases. [ 12 points] Solution. The relationship between the partition function and the average energy is given in equation (17.20) and (17.21) on p. 698 of the textbook h E i =- ∂ ln Q ∂β N,V = k B T 2 ∂ ln Q ∂T N,V (5) so applying (1) and using the property of logarithms that log ab = log a + log b : h E i = k B T 2 ∂ ∂T 2 X i =1 N i ln q i ( V,T )- ln N i ! ! (6) = 2 X i =1 N i k B T 2 ∂ ∂T ln q i ( V,T ) (7) Note that ln N i ! are constants and therefore vanish under differentiation. Now applying (2) and using the properties of logs: h E i = 2 X i =1 N i k B T 2 3 2 ∂ ∂T " ln T + ln 2 mk B h 2 3 2 V # (8) = 2 X i =1 N i k B T 2 3 2 1 T + 0 = 3 2 k B T 2 X i =1 N i = 3 2 ( N 1 + N 2 ) k B T (9) 1 CHEM 444 HW#3 Solutions The relationship between the partition function and the pressure is given by equation (17.32) p. 705 of the textbook: h P i = k B T ∂ ln Q ∂V N,T (10) as above, applying (1) and using the property of logarithms that log ab = log a + log b gives: h P i = k B T 2 ∂ ∂V 2 X i =1 N i ln q i ( V,T )- ln N i ! ! (11) = 2 X i =1 N i k B T 2 ∂ ∂V ln q i ( V,T ) (12) = 2 X i =1 N i k B T 2 ∂ ∂V " ln 2 mk B T h 2 3 2 + ln V # (13) = 2 X i =1 N i k B T 2 V = ( N 1 + N 2 ) k B T 2 V (14) Postulating that the ensemble average pressure is equal to the observable pressure, P = h P i (15) therefore: PV = h P i V = ( N 1 + N 2 ) k B T 2 V · V = ( N 1 + N 2 ) k B T 2 (16) Note. I want to emphasise that the connection between the ensemble average and a physical observable is a postulate that must be assumed to be true. Note on approximations. Several people employed Stirling’s approximation to simplify ln N !. Firstly, Stirling’s approximation is an approximation and should be denoted as such, i.e ln N ! ≈ N ln N- N (17) not as an equality ln N ! = N ln N- N (18) (other symbols for approximation, e.g. ' , ∼ = w but not ∼ are acceptable. ∼ denotes a crude approx- imation.) Secondly, Stirling’s approximation is not necessary to derive the results. Employing unnecessary approximations reduce the generality of the final result. In the future, this may be treated as aapproximations reduce the generality of the final result....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

sol3 - CHEM 444 HW#3 Answer key by Jiahao Chen due Grading...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online