Which of the following thermodynamic identities is INCORRECT?

Question

Professor PC Thomas Classes · Answer

Let's analyze each thermodynamic identity:
A) The heat absorbed at constant pressure is equal to the change in enthalpy, i.e., $d q_P=d H$.

According to the first law of thermodynamics, $d U=d q+d W$. For a process at constant pressure, if only $\mathrm{P}-\mathrm{V}$ work is considered, $d W=-P d V$. So, $d U=d q-P d V$. Rearranging, $d q=d U+P d V$. Enthalpy $(H)$ is defined as $H=U+P V$.
Differentiating $H: d H=d U+P d V+V d P$. At constant pressure $(d P=0), d H=d U+P d V$. Comparing this with $d q_P=d U+P d V$, we find $d q_P=d H$.
This statement is correct.
B) The heat absorbed at constant volume is equal to the change in internal energy, i.e., $d q_V=d U$.

According to the first law of thermodynamics, $d U=d q+d W$.
For a process at constant volume ( $\boldsymbol{d} \boldsymbol{V}=\mathbf{0}$ ), the work done $\boldsymbol{d} \boldsymbol{W}=-\boldsymbol{P} \boldsymbol{d} \boldsymbol{V}=\mathbf{0}$. Therefore, $\boldsymbol{d} \boldsymbol{U}=\boldsymbol{d} q_{\boldsymbol{V}}$. This statement is correct.
C) For a system at constant temperature, the change in Gibbs free energy with pressure is given by $\left(\frac{\partial G}{\partial P}ight)_T=-V$.

The Gibbs free energy $(\boldsymbol{G})$ is defined as $\boldsymbol{G}=\boldsymbol{H}-\boldsymbol{T} \boldsymbol{S}$. We also know that $\boldsymbol{H}=\boldsymbol{U}+\boldsymbol{P} \boldsymbol{V}$.
Substituting $H$ into the Gibbs free energy definition: $G=U+P V-T S$.
To find the differential of $\boldsymbol{G}$, we differentiate this expression:
$$
d G=d U+P d V+V d P-T d S-S d T
$$

For a reversible process, from the first law of thermodynamics, $d U=d q_{	ext {rev }}+d W_{	ext {rev }}$.
For P-V work, $d W_{	ext {rev }}=-P d V$, and for reversible heat transfer, $d q_{	ext {rev }}=T d S$.
So, $\boldsymbol{d} U=\boldsymbol{T} \boldsymbol{d} \boldsymbol{S}-\boldsymbol{P} \boldsymbol{d} \boldsymbol{V}$.
Substitute this expression for $d U$ into the $d G$ equation:
$$
\begin{aligned}
& d G=(T d S-P d V)+P d V+V d P-T d S-S d T \
& d G=V d P-S d T
\end{aligned}
$$

From this fundamental equation for $\boldsymbol{d} \boldsymbol{G}$, we can identify the partial derivatives: \ $\left(\frac{\partial G}{\partial P}ight)_T=V$ (when temperature $T$ is constant, $d T=0$ ).
$\left(\frac{\partial G}{\partial T}ight)_P=-S$ (when pressure $P$ is constant, $d P=0$ ).
The given identity is $\left(\frac{\partial G}{\partial P}ight)_T=-V$. This statement is incorrect.
D) The molar heat capacity at constant pressure is defined as $C_P=\left(\frac{\partial H}{\partial T}ight)_P$.

Heat capacity $(C)$ is generally defined as the amount of heat required to raise the temperature by one degree, $C=\frac{d q}{d T}$. At constant pressure, we know $d q_P=d H$.
Therefore, the heat capacity at constant pressure is $C_P=\left(\frac{\partial H}{\partial T}ight)_P$.
This statement is correct.
Therefore, the incorrect thermodynamic identity is C .