HW13_Sol_F10

# HW13_Sol_F10 - Nov 29 2010 PHYS 2101 HW#13 WileyPlus...

This preview shows pages 1–2. Sign up to view the full content.

Nov. 29, 2010 PHYS 2101 HW#13 WileyPlus Problem Solutions 1 HW #13 Solutions – or at least most of them… (USING BOOK VALUES 9 th Edition in problems; problem # indicated by []) 1) [19-CQ2] The change in the internal energy ( Δ E int ) only depends on the endpoints (i.e. it’s path independent), thus Δ E ac = Δ E abc = Δ E ab + Δ E bc ( ) . Since the path ab is isothermal, Δ E ab = 0 (because Δ E ab = Q ab W ab , here W ab = Q ab = 5 J ). Moreover, since the path bc is adiabatic, Q bc = 0 and because Δ E bc = Q bc W bc , here Δ E bc = W bc . Putting this together Δ E ac = Δ E ab + Δ E bc ( ) = 0 J + ( 4 J) ( ) = 4 J . 2) [19-2] . (a) Equation 19-3 yields n = M sam / M = 2.5/197 = 0.0127 mol. (b) The number of atoms is found from Eq. 19-2: N = nN A = (0.0127)(6.02 × 10 23 ) = 7.64 × 10 21 . 3) [19-13] . (a) At point a , we know enough information to compute n : (b) We can use the answer to part (a) with the new values of pressure and volume, and solve the ideal gas law for the new temperature, or we could set up the gas law in terms of ratios (note: n a = n b and cancels out): which yields an absolute temperature at b of T b = 1.8 × 10 3 K. (c) As in the previous part, we choose to approach this using the gas law in ratio form: which yields an absolute temperature at c of T c = 6.0 × 10 2 K.

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.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern