I am trying to derive certain formulas and and I feel like the following expression is good for my derivation. $$ \left ( \frac{\partial u}{\partial T} \right )_p= \left ( \frac{\partial u}{\partial T} \right )_V $$ I am trying to say that the above expression is true because of first law of thermodynamics. any suggestions weather if this approach is good or not
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$\begingroup$ What's the equation for $u$ in terms of $T, p$ and $V$? $\endgroup$– SuperCiociaCommented Mar 3, 2019 at 17:36
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$$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$ $$dV=\left(\frac{\partial V}{\partial T}\right)_PdT+\left(\frac{\partial V}{\partial P}\right)_TdP$$so
$$dU=\left[\left(\frac{\partial U}{\partial T}\right)_V+\left(\frac{\partial U}{\partial V}\right)_T\left(\frac{\partial V}{\partial T}\right)_P\right]dT+\left(\frac{\partial U}{\partial V}\right)_T\left(\frac{\partial V}{\partial P}\right)_TdP$$ So, $$\left(\frac{\partial U}{\partial T}\right)_P=\left[\left(\frac{\partial U}{\partial T}\right)_V+\left(\frac{\partial U}{\partial V}\right)_T\left(\frac{\partial V}{\partial T}\right)_P\right]$$ According to this, your equation is correct for an ideal gas, but not for a real gas beyond the ideal gas region.
answered Mar 3, 2019 at 17:57
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$\begingroup$ @VincentFraticelli Your equation for Cp is incorrect. Cp is the partial derivative of enthalpy H with respect to temperature at constant pressure (not the partial derivative of internal energy U with respect to temperature at constant pressure). And, who said Cp and Cv are equal? $\endgroup$ Commented Mar 3, 2019 at 20:43
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$\begingroup$ OK ! I had a moment of confusion ! $\endgroup$ Commented Mar 4, 2019 at 5:45