# Chemical Forums

## Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Il Divo on January 04, 2014, 01:42:44 PM

Title: Work Equation and Entropy
Post by: Il Divo on January 04, 2014, 01:42:44 PM
Hey all,

just reviewing my Atkins Thermodynamics and Kinetics text and had some different points that I'm pretty stuck on.

1) My first issue relates to the rules of Calculus, with regard to expansion work. dw = -p (external) dv. The text explains that if pressure is constant then we may bring it outside the integral, so w = -p (external) ΔV. But before we establish that the pressure is constant, shouldn't the equation read as dw = -dp (external) dV?

2) With regard to the above, why is it that we are able to say dw or dq, if these are path functions? If "d" indicates change and work and heat are both considered processes, what exactly does it mean to say dw or dq? With Temperature, it seems straight forward. T indicates the value of a property, while dT would indicate a change in that property.

3) My textbook explains that a change in entropy can be regarded as the heat supplied reversible over temperature. So ds = dq rev/ T. End result is that for a reversible process, there should be no increase in the entropy of the universe, assuming there is thermal equilibrium. What I'm having trouble understanding is why is entropy defined as heat supplied reversibly? The text explains that in determining entropy, you simply find a reversible path between them, but assuming that the process is irreversible, the actual heat being supplied will be substantially different from the heat assumed by a reversible pathway, wouldn't it?

Thanks in advance. I apologize if I'm missing something really basic here. I find when reading Atkins it's easy to lose myself in the language of the mathematics, making it difficult to keep track of what's actually happening.
Title: Re: Work Equation and Entropy
Post by: orgo814 on January 04, 2014, 04:55:54 PM
I only looked at point one. But, for work the equation is w = -p dv. For ideal gas, p = nrt/V.. in other words nrt x 1/V. We're integrating in terms of V so we can remove anything unnecessary for the integration (NRT). so w = nrt integ (1/V). The rules of calculus state that 1/V is equal to ln V (ln Vf/Vi). Combining all these factors, w = -nrt ln Vf/Vi. For real gases, the integration will be a little more difficult but it follows the same idea. Hope that helped