Ps10soln - PHYSICS 218 SOLUTION TO HW 10 Created 23:26pm Last updated December 9 2004 1:01am 1 Schroeder 3.32(a In moving the piston through a

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PHYSICS 218 SOLUTION TO HW 10 Created: November 25, 2004 23:26pm Last updated: December 9, 2004 1:01am 1. Schroeder 3.32 (a) In moving the piston through a distance of Δ x = 1 mm, the work done on the system is W = F Δ x = 2000 × 10 - 3 = 2 J. (b) We recall that a fast compression is an adiabatic process. Therefore, no heat is added to the gas. (c) Assuming that all the work done on the system is converted into the internal energy of the gas, the increase of internal energy is therefore Δ U = 2 J. (d) Calculating the change in entropy appears to be complicated because all of our familiar thermody- namic variables change in the process. That is, P f , V f , U f , T f are all different from P i , V i , U i , T i . As is always the case, there are several ways to calculate the change in entropy. Schroeder suggests that you use the thermodynamic identity d U = T d S - P d V to calculate the change in entropy. In this case, we want to look at the process in the U - V plane as outlined on pp. 111-112 of Schroeder. Our process is one in which U i < U f and V i > V f , so the two can be represented by two points in a figure like Fig. (3.15), except that – contrary to the figure – V f is below V i in our case. We envision replacing the actual process by two quasistatic processes, in which Step 1 is an increase in U from U i to U f with V held constant (as in Fig. 3.15) and Step 2 is a compression or step downward from V i to V f with U held constant at U i (just opposite to Step 2 in Fig. 3.15). Since V is constant in Step 1, we have d U = T d S and Δ S 1 = Z d U T = C V Z d T T = C V ln ± T f T i ² . Now, T f = T i + Δ U/C V , so ln( T f /T i ) = ln[1 + Δ U/ ( C V T i )] Δ U/ ( C V T i ) and Δ S 1 Δ U/T i = W/T i . During Step 2, we have T d S = P d V since U (ergo T ) is constant. Hence, we have Δ S 2 = (1 /T ) R P d V , so Δ S 2 = NkT T Z d V V = Nk ln ± V f V i ² = Nk ln ³ 1 + Δ V V i ´ Nk Δ V V i . (Note Δ V < 0!). Since Nk/V i = P i /T i , we have Δ S 2 P i Δ V/T i . Putting these together we find that the total change in entropy Δ S is, Δ S = Δ S 1 + Δ S 2 ( W + P i Δ V ) /T i . Then, Δ V = - A Δ x = - 10 - 2 m 2 × 10 - 3 m = - 10 - 5 m 3 , so P i Δ V ≈ - 1 J. Finally, Δ S (2 J - 1 J) / (300 K) = 3 . 33 × 10 - 3 J/K. This is positive as it should be since the process that is got us from i to f is sudden – anything but quasistatic. It is possible to calculate Δ S in the more familiar P - V plane. In this case you also need to define a third point in order to get from i to f along a quasistatic path. You could use a quasistatic isothermal or a quasistatic adiabatic process along with a quasistatic process at constant volume. You could also use both an isothermal and an adiabatic process. Since any of these choices require calculating P and V at this third point, they turn out to be more complicated analytically than the calculation in the U - V plane where quasistatic paths with one variable held constant are simple. 2.
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This note was uploaded on 09/28/2008 for the course PHYS 218 taught by Professor Pollack during the Fall '04 term at Cornell University (Engineering School).

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Ps10soln - PHYSICS 218 SOLUTION TO HW 10 Created 23:26pm Last updated December 9 2004 1:01am 1 Schroeder 3.32(a In moving the piston through a

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