HW13_Sol_F10

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

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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. (d) The net energy added to the gas (as heat) is equal to the net work that is done as it progresses through
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This note was uploaded on 01/26/2011 for the course PHYS 2101 taught by Professor Grouptest during the Spring '07 term at LSU.

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

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