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

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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.

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

1 Answer 1


If we consider the molar volume $V$ as a function of pressure and temperature, i.e. $V = V(p,T)$ we have \begin{equation} \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 \end{equation} 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}


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