# equation of state for a liquid in small ranges of temperature and pressure

consider the following question from Physical chemistry by Robert J. Silbey Robert A. Alberty Moungi G. Bawendi

the answer is given to be $$V=Ke^{\alpha T}e^{-kp}$$

But I do not understand how this was derived. Using the definition of $$k$$ and $$\alpha$$ I got the individual parts of the equation as $$V=V_{0}e^{\alpha T}$$ and $$V=V_{0}e^{-k p}$$ But I do not know how to combine them as first one is with constant pressure and second has constant temprature. Also nothing is mentioned about the variable/constant (I don't know what it is) K.

• $K=V(T_0) \cdot \exp{(- \alpha \cdot T_0)} \cdot \exp{(k \cdot p_0})$ Apr 3 at 10:32
• Should be V(T0,p0) Apr 3 at 10:42

If we consider the molar volume $$V$$ as a function of pressure and temperature, i.e. $$V = V(p,T)$$ we have $$$$\mathrm{d}V = \left(\frac{\partial V}{\partial p}\right)_T\mathrm{d}p + \left(\frac{\partial V}{\partial T}\right)_p\mathrm{d}T \tag1$$$$ Dividing Eq. (1) by $$V$$ on both sides of Eq. (1) makes the isobaric expansivity $$\beta$$ and isothermal compressibility $$\kappa$$ appear. If we consider them constant, we can integrate on both sides and get \begin{align} \frac{\mathrm{d}V}{V} &= \frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_T\mathrm{d}p + \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p\mathrm{d}T \\ \frac{\mathrm{d}V}{V} &= -\kappa \mathrm{d}p + \beta \mathrm{d}T \\ \int_{V_0}^{V} \frac{\mathrm{d}V}{V} &= -\kappa \int_{p_0}^p \mathrm{d}p + \beta \int_{T_0}^T \mathrm{d}T \\ \ln\left(\frac{V}{V_0}\right) &= -\kappa (p - p_0) + \beta (T - T_0) \\ V &= V_0 \exp\left[-\kappa (p - p_0) + \beta (T - T_0)\right] \tag3 \end{align} and to obtain the result from the book we apply the exponentiation properties \begin{align} V &= \underbrace{V_0\exp(\kappa p_0) \exp(-\beta T_0)}_{\equiv K} \exp(-\kappa p) \exp(\beta T) \rightarrow \boxed{V = K e^{-\kappa p} e^{\beta T}} \tag3 \end{align}