P21_065 - i → a segment o± path 2 Hence the integrand f...

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65. First we show that R dQ is path-dependent. To do this all we need is to show that R dQ is diferent ±or at least two separate paths, say path 1 and 2, as depicted in the ²gure below. We write R dQ = R pdV + R nC V dT . The second term on the right, R nC V dT , yields nC V T upon integration and is obviously path-independent. The ²rst term, R pdV , however, is diferent ±or the two paths. In ±act R f i pdV along path 1 is greater than that along path 2, by the area o± the shaded triangle enclosed by the two paths. There±ore, R dQ is indeed path-dependent. f 2 2 a i 1 V p Now we consider R TdQ = R pTdV + R nC V TdT . Once again the second term on the right, R nC V TdT , yields 1 2 nC V T 2 upon integration and is path-independent. The ²rst term, R pTdV , however, yields a higher value along path 1 than path 2. To see that, note that Z 2 pT dV = Z i a pT dV + Z a f pT dV = Z i a pT dV . Now, i± we compare the two integrals, R 1 pTdV and R i a pT dV , we realize that the average values o± both T and p along path 1 are greater than their respective corresponding values along the
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Unformatted text preview: i → a segment o± path 2. Hence, the integrand f ( p, T ) = pT is always greater along path 1. Thus, the two integrals over V , which have the same upper and lower limits, are not equal to each other: Z 1 pT dV > Z i → a pT dV = Z 2 pT dV . We see then that R TdQ is greater along path 1 than path 2 and is there±ore path-dependent. Similarly, one can show that ±or R dQ/T 2 = R pdV/T 2 + R nC V dT/T 2 , the second term on the right is path-independent, while ±or the ²rst term Z pdV/T 2 = nR Z dV TV , we have nR Z 2 dV TV = nR Z i → a dV TV > nR Z 1 dV TV , since the average value o± 1 /T is greater along along the i → a segment o± path 2 than on path 1. Consequently, R dQ/T 2 is less along path 1 than path 2 and is there±ore path-dependent....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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