Chapter5_B

# Chapter5_B - Chapter 5 Section B Non-Numerical Solutions...

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Chapter 5 - Section B - Non-Numerical Solutions 5.1 Shown to the right is a PV diagram with two adi- abatic lines 1 2 and 2 3, assumed to inter- sect at point 2. A cycle is formed by an isothermal line from 3 1. An engine traversing this cycle would produce work. For the cycle U = 0, and therefore by the first law, Q + W = 0. Since W is negative, Q must be positive, indicating that heat is absorbed by the system. The net result is therefore a complete conversion of heat taken in by a cyclic process into work, in violation of Statement 1a of the second law (Pg. 160). The assumption of intersecting adiabatic lines is there- fore false. 5.5 The energy balance for the over-all process is written: Q = U t + E K + E P Assuming the egg is not scrambled in the process, its internal-energy change after it returns to its initial temperature is zero. So too is its change in kinetic energy. The potential-energy change, however, is negative, and by the preceding equation, so is Q . Thus heat is transferred to the surroundings. The total entropy change of the process is: S total = S t + S t surr Just as U t for the egg is zero, so is S t . Therefore, S total = S t surr = Q surr T σ = Q T σ Since Q is negative, S total is positive, and the process is irreversible. 5.6 By Eq. (5.8) the thermal efficiency of a Carnot engine is: η = 1 T C T H Differentiate: ∂η T C T H = − 1 T H and ∂η T H T C = T C T H 2 = T C T H 1 T H Since T C / T H is less unity, the efficiency changes more rapidly with T C than with T H . So in theory it is more effective to decrease T C . In practice, however, T C is fixed by the environment, and is not subject to control. The practical way to increase η is to increase T H . Of course, there are limits to this too. 5.11 For an ideal gas with constant heat capacities, and for the changes T 1 T 2 and P 1 P 2 , Eq. (5.14) can be rewritten as: S = C P ln T 2 T 1 R ln P 2 P 1 ( a ) If P 2 = P 1 , S P = C P ln T 2 T 1 If V 2 = V 1 , P 2 P 1 = T 2 T 1 Whence, S V = C P ln T 2 T 1 R ln T 2 T 1 = C V ln T 2 T 1 642

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Since C P > C V , this demonstrates that S P > S V . ( b ) If T 2 = T 1 , S T = − R ln P 2 P 1 If V 2 = V 1 , T 2 T 1 = P 2 P 1 Whence, S V = C P ln P 2 P 1 R ln P 2 P 1 = C V ln P 2 P 1 This demonstrates that the signs for S T and S V are opposite. 5.12 Start with the equation just preceding Eq. (5.14) on p. 170: dS R = C ig P R dT T d ln P = C ig P R dT T d P P For an ideal gas PV = RT , and ln P + ln V = ln R + ln T . Therefore, d P P + dV V = dT T or d P P = dT T dV V Whence, dS R = C ig P R dT T dT T + dV V = C ig P R 1 dT T + d ln V Because ( C ig P / R ) 1 = C ig V / R , this reduces to: dS R = C ig V R dT T + d ln V Integration yields: S R = T T 0 C ig V R dT T + ln V V 0 * * * * * * * * * * * * * * * * * * * * * * As an additional part of the problem, one could ask for the following proof, valid for constant heat capacities. Return to the original equation and substitute dT / T = d P / P + dV / V : dS R = C ig P R d P P + C ig P R dV V d P P = C ig V R d P P + C ig P R dV V Integration yields: S R = C ig V R ln P P 0 + C ig P R ln V V 0 5.13 As indicated in the problem statement the basic differential equations are: dW dQ H dQ C = 0 ( A ) dQ H dQ C = − T H
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