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One-dimensional polymer (Statistical Physics)
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[QUOTE="CharlieCW, post: 6150422, member: 649017"] Indeed, the term ##pV_S## is for the pressure and volume, but since the general formula was derived for a 3D recipient I was thinking about converting it to the one-dimensional case ##pV_S\rightarrow \tau L##. However, it also makes more sense that you mention to obtain the tension as ##dZ/dl##, as in that way I could relate the tension to the length of the chain. I also checked how to solve the sums by approximating some series so I think I should be fine once I know how to write the partition function. However, I'm still unsure on how to express the partition function, that is, how to write the exponential in terms of the length and energy of the molecule. Here's my guess (correct me if I'm wrong). From the statement, if I have ##n## molecules with length ##l_1##, then I have ##N-n## molecules with length ##l_2##, so that means that the total energy for a given state ##\{ n \}## should be ##E_n=nE_1+(N-n)E_2=n(E_1-E_2)+NE_2##, and thus, $$exp[-\beta E_n]=exp[-\beta n(E_1-E_2)+NE_2]$$ Now, how could I include the length of the chain ##L##? [/QUOTE]
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One-dimensional polymer (Statistical Physics)
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