ãáÊÞì ÇáÝíÒíÇÆííä ÇáÚÑÈ > ÞÓã ÇáãäÇåÌ ÇáÏÑÇÓíÉ > ÝíÒíÇÁ ÇáãÑÍáÉ ÇáÌÇãÚíÉ. | ||
ÓÄÇá Ýí ÇáÝíÒíÇÁ ÇáÇÍÕÇÆíå æÇÍÊÇÌ ÌæÇÈ ãä ÝÖáßã |
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ÃÏæÇÊ ÇáãæÖæÚ | ÇäæÇÚ ÚÑÖ ÇáãæÖæÚ |
#1
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![]() ÈÓã Çááå ÇáÑÍãä ÇáÑÍíã
ÇáÓáÇã Úáíßã æÑÍãÉ Çááå æÈÑßÇÊå Çááå íæÝÞßã ÇÐÇ ÇÍÏ íÚÑÝ ÇÓÊäÊÇÌ ÇáÍÏ ÇáÇÞÕì ááÇäÊÑæÈí ÇáÇÍÕÇÆí Smax=KB log (M ![]() ãÍÊÇÌÊå ÖÑæÑí Çááå áÇíåíäßã æ áßã ãäí ÃÌãá ÊÍíÉ . |
#2
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![]() ÇáÝÒÚå Çááå íÌÒÇßã ÇáÌäå
ãÇÙäí Çäßã ÊÑÏæäí ÎÇíÈå æÇãáí ÈÇááå Ëã Èßã ßÈíÑ Úáì Çãá ÇáÑÏ ÞÑíÈÇ ÊÍíÇÊí |
#3
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![]() ÇáÓáÇã Úáíßã æÑÍãÉ Çááå æÈÑßÇÊå
what do you mean by KB &M? If they mean : Kb is boltzmann constant M is thermodynamic probabilty I know the the derivation let me know |
#4
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![]() ÇæáÇ äÔßÑßã Úáì ÇáÑÏ
ÈÓ ÇäÇ ÇÈÛì ÇÓÊäÊÇÌ åÐÇ ÇáÞÇäæä (ßíÝ íßæä) ÇÐÇ ÊÚÑÝå ¿ æáßã ÎÇáÕ ÊÍíÇÊí |
#5
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ÑÏ: ÓÄÇá Ýí ÇáÝíÒíÇÁ ÇáÇÍÕÇÆíå æÇÍÊÇÌ ÌæÇÈ ãä ÝÖáßã
ÇáÓáÇã Úáíßã æÑÍãÉ Çááå æÈÑßÇÊå
The total entropy S of two independent systems given by S = S1 +S2 , on other hand the thermodynamics of the system is M =M1*M2 Assume S is unknown function of M ; S= J(M) in such way: J(M1) + J(M2) = J (M1*M2) .................................[1] I. Take partial derivative of equation [1] with respect to M1( dJ(M1))/dM1 = (dJ(M1*M2))/(d(M1*M2))*(d(M1*M2))/dM1 II. Take partial derivative of equation [1] with respect to M2(dJ(M1))/dM1 = M2 * (dJ(M1*M2))/(d(M1*M2)) ……………………………….[2] (dJ(M2))/dM2 = (dJ(M1*M2))/(d(M1*M2))*(d(M1*M2))/dM2 From [2] &[3] (dJ(M2))/dM2 = M1 * (dJ(M1*M2))/(d(M1*M2)) ……………………………….[3] M1 (dJ(M1))/dM1 = M2 (dJ(M2))/dM2 ………………………………….[4] Each term in 4 equal constant (KB) KB=M (dJ(M))/dM [align=center]dJ(M)= KB dM/M ; Integrate both side J(M) = KB log (M) S = KB log (M) [/align] :laughter01: |
#6
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![]() æÚáíßã ÇáÓáÇã æÑÍãÉ Çááå æÈÑßÇÊå
æÇáÔßÑ ÇáÌÒíá áß Îíæ¡ ãæÕæá ÈÇáÏÚÇÁ áß æÇÔåÏ Çäß æÇááå ãÇÞÕÑÊ ÌÇÊ ÇáÇÌÇÈå Ýí æÞÊåÇ ÌÚáåÇ Çááå Ýí ãæÇÒíä ÍÓäÇÊß ÊÍíÇÊí ãÚ ÇáÔßÑ |
ÇáÐíä íÔÇåÏæä ãÍÊæì ÇáãæÖæÚ ÇáÂä : 1 ( ÇáÃÚÖÇÁ 0 æÇáÒæÇÑ 1) | |
ÇäæÇÚ ÚÑÖ ÇáãæÖæÚ |
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