Unformatted text preview: μ = . We can integrate to solve for F , and we obtain: F = Nτ log 1 The Pressure of an Ideal Gas We seek to get the pressure from the free energy. This is no problem though, since we can recall or rederive that p =  . Looking at the expression for F above, we see that we can expand it to be the sum of many terms, most of which have no V dependence. The derivative becomes simple, and returns something familiar: p = This is the ideal gas law. If it doesn't look familiar, recall that the chemistry version uses number of moles instead of number of particles, and replaces the temperature as we've defined it with the temperature in Kelvin. You might wish to work out the conversion to assure yourself....
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This note was uploaded on 02/09/2012 for the course PHY PHY2053 taught by Professor Davidjudd during the Fall '10 term at Broward College.
 Fall '10
 DavidJudd
 Physics

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