Problem NO. 1 : Law of the Adiabatic Process


Reading R. Resnick – D. Haliday thermodynamics, I’ve found something well known me although it’s making me wonder every single time when I think about it. It’s the derivation of the thermodynamics law p V^{\kappa} = const that goes like this:
From first law of thermodynamics \Delta Q = \Delta U + \Delta W , but in the adiabatic process we have \Delta Q = 0 and instead of \Delta W we put p\Delta V

And this is the moment that I do not understand! Why, to hell do we put p\Delta V instead of p\Delta V + \Delta pV if both of the variables DO CHANGE in time?! Even if it’s in diffrential, it should be that second one.

If someone knows the reasons why it goes like that, please comment this post and explain it to me.

Futher we get 0 = c_v n \Delta T + p \Delta V because \Delta U = c_v n \Delta T , and from it \Delta T = - \frac{p \Delta V}{n c_v}
On the other hand from the perfect gas equatation pV = nRT brought to the diffrental form we get \Delta T = \frac{p \Delta V + V \delta p}{nR} After comparing those two expressions for \Delta T , remembering that C_p - C_v = R and \kappa = \frac{c_p}{c_v} and some simple algebraic transformation, we recieve \frac{dp}{p} + \kappa \cdot \frac{dV}{V} = 0
Integrating gives us finnaly \ln{p} + \kappa \cdot \ln{V} = const , hence pV^{\kappa} = const

By the way I’ve noticed that there exist a familiar strange derivation in mathematics, for derivative from product of two functions, maybe because it’s just a simplified derivation of something more complicated.

\lim\limits_{\Delta x\rightarrow0}\frac{\left[fg\right](x+\Delta x)-\left[fg\right](x)}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}

Now the trick we’ll use to evaluate this limit is to add and subtract f(x+\Delta x) \cdot g(x)  to the numerator here. That is, in effect we’re adding zero and leaving it alone, but the formula will be easier to work with. In particular, we can start splitting it up using the laws of limits.

\lim\limits_{\Delta x\rightarrow0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x)+f(x+\Delta x)g(x)-f(x)g(x)}{\Delta x}= \newline \left(\lim\limits_{\Delta x\rightarrow0}f(x+\Delta x)\right)\left(\lim\limits_{\Delta x\rightarrow0}\frac{g(x+\Delta x)-g(x)}{\Delta x}\right)+\left(\lim\limits_{\Delta x\rightarrow0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\right)\left(\lim\limits_{\Delta x\rightarrow0}g(x)\right)

Of these four limits, the fourth is the limit of a continuous function because g(x) doesn’t depend on \Delta x . The second and third are just the definitions of f'(x) and g'(x) . The first limit goes to f(x) because, as we showed, all differentiable functions are continuous. And so we have the rule



2 Responses to “Problem NO. 1 : Law of the Adiabatic Process”

  1. 1 Catur B.S.

    I have answer in why we put p delta v instead of delta p v plus p delta v.

    From the definition of work in translasional motion :
    The work done by an agent exerting a constant force (F) and causing a displacement (s) equals the magnitude of the displacement, s, times the component of F along the direction of s.

    In Fluida the formula is p delta V, because pressure p multiply by area A will be F. and displacement along direction of F along delta v = s.

  2. Thank you for your answer Catur B.S! However here you are the first there were some guys on IRC physics channel who explained it to me fimillary – that it’s due to Work definition.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: