Recent content by Dave Mata

I tried to verify your equation but was unsuccessful. Here is my process:

The values that were given as a part of the problem (I did not include these in the original post of this thread as I was struggling more with the appropriate derivation) were $$c_P = 24 \frac{J}{mol K}, β=6*10^-7 atm^-1, α=15*10^-6 K^-1 $$ Regarding your equation to approximate 1/T, is it...

Using your approximation, I get ΔT=0.134 C. Thanks for the explanation. I guess I had been making it more complicated by doing a transform of dS when I didn't need to. For the integration of (4), it is the LHS I'm having trouble with. I split the exponential so I was left with $$(C_P e^{α T_0})...

If we can go back to the integration, I'm not sure how to get to your answer. The equation I'm integrating is $$dS=(\frac{C_P}{T}-\frac{Vα^2}{β})dT+\frac{α}{β} dV$$ I plug in the equation $$V=V_0 exp(α ΔT-β ΔP)$$ And set dS=0 $$0=(\frac{C_P}{T}-\frac{(V_0 exp(α ΔT-β ΔP))α^2}{β})dT+\frac{α}{β}...

After plugging in and integrating (and some algebraic simplification) I get the following $$c_P*ln(\frac{T_2}{T_1}) +V_0 \frac{α}{β} (exp(αΔT-βΔP) (1-αΔT)-1)$$ Now all the quantities in my equation are known (Cp, T1, α, β, ΔP), and I am solving for T2. When you say specific volume, I'm assuming...

Here is my best guess: $$dV=Vα dT-Vβ dP$$ If that is what you meant, I see that I can solve for dT and integrate. If I do, I get: $$ΔT=\frac{1}{α} ln(\frac{V_2}{V_1})+\frac{β}{α} ΔP$$ However, this doesn't seem to be the answer that I'm supposed to get, since the second part of the problem only...

Homework Statement Derive an expression for the change of temperature of a solid material that is compressed adiabatically and reversible in terms of physical quantities. (The second part of this problem is: The pressure on a block of iron is increased by 1000 atm adiabatically and...