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Topic: Internal Energy Question  (Read 2745 times)

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Offline G O D I V A

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Internal Energy Question
« on: October 18, 2009, 08:24:41 AM »
I dont understand how my book got this:

It says when V changes to V + dV at constant temperature, U changes to (dU/dV)TdV; and when T changes to T + dT at constant volume, U changes to (dU/dT)VdT, thus the overall equation is:

dU = (dU/dV)TdV + (dU/dT)VdT

I dont understand what they are differentiating to get this
« Last Edit: October 18, 2009, 08:41:23 AM by G O D I V A »

Offline Yggdrasil

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Re: Internal Energy Question
« Reply #1 on: October 18, 2009, 02:03:12 PM »
So, we would like to find the change in internal energy (dU) when we increase the volume from V to V+dV.  Let's setup an equation for this:

dU = U(V+dV,T) - U(V,T)

Now, how do we calculate U(V+dV,T)?  Because dV is small, let's assume that the function is a straight line near this point, so to find U(V+dV,T), we model it as a line (ignoring temperature):

U(V+x) = U(V) + m*x

where m is the slope.  Here, the slope is just the partial derivative of the internal energy with respect to V: (dU/dV)T.  So, this tells us that;

U(V+dV,T) = U(V,T) + (dU/dV)TdV

Now, we plug this back into our equation for dU above:

dU = U(V+dV,T) - U(V,T) = U(V,T) + (dU/dV)TdV - U(V,T) = (dU/dV)TdV

You can go through a similar line of reasoning to find dU when T is increased by dT.

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