Skip to main content
Chemistry LibreTexts

10.1: Thermodynamic Relationships from dE, dH, dA and dG

  • Page ID
  • In Chapter 9, we substitute \(dq^{rev} = TdS\), from the second law, into

    \[dE = dq + PdV\]

    from the first law, to obtain, for any closed system undergoing a reversible change in which the only work is pressure–volume work, the fundamental equation, \(dE = TdS + PdV\). In view of the mathematical properties of state functions that we develop in Chapter 7, this result means that we can express the energy of the system as a function of entropy and volume, \(E = E\left(S,V\right)\). With this choice of independent variables, the total differential of \(E\) is

    \[(dE = {\left({\partial E}/{\partial S}\right)}_VdS + {\left({\partial E}/{\partial V}\right)}_SdV.\]

    Equating these expressions for \(dE\), we find

    \[\left[{\left(\frac{\partial E}{\partial S}\right)}_V + T\right]dS + \left[{\left(\frac{\partial E}{\partial V}\right)}_S + P\right]dV\mathrm{=0}\]

    for any such system. Since \(S\) and \(V\) are independent variables, this equation can be true for any arbitrary state of the system only if the coefficients of\(\mathrm{\ }dS\) and \(dV\) are each identically equal to zero. It follows that

    \[{\left(\frac{\partial E}{\partial S}\right)}_V = T\]


    \[{\left(\frac{\partial E}{\partial V}\right)}_S\mathrm{=-}P\]

    Moreover, because dE is an exact differential, we have

    \[\frac{\partial }{\partial V}{\left(\frac{\partial E}{\partial S}\right)}_V = \frac{\partial }{\partial S}{\left(\frac{\partial E}{\partial V}\right)}_S\]

    so that

    \[{\left(\frac{\partial T}{\partial V}\right)}_S\mathrm{=-}{\left(\frac{\partial P}{\partial S}\right)}_V\]

    Using the result \(dH = TdS + VdP\), parallel arguments show that enthalpy can be expressed as a function of entropy and pressure, \(H = H\left(S,P\right)\), so that

    \[{\left(\frac{\partial H}{\partial S}\right)}_P = T\]


    \[{\left(\frac{\partial H}{\partial P}\right)}_S = V\] and \[{\left(\frac{\partial T}{\partial P}\right)}_S = {\left(\frac{\partial V}{\partial S}\right)}_P\]

    Since \(dA\mathrm{=-}SdT + PdV\), the Helmholtz free energy must be a function of temperature and volume, \(A = A\left(T,V\right)\), and we have

    \[{\left(\frac{\partial A}{\partial T}\right)}_V\mathrm{=-}S\] and \[{\left(\frac{\partial A}{\partial V}\right)}_T\mathrm{=-}P\] and \[{\left(\frac{\partial S}{\partial V}\right)}_T = {\left(\frac{\partial P}{\partial T}\right)}_V\]

    Likewise, \(dG\mathrm{=-}SdT + VdP\) implies that the Gibbs free energy is a function of temperature and pressure, \(G = G\left(P,T\right)\), so that

    \[{\left(\frac{\partial G}{\partial T}\right)}_P\mathrm{=-}S\]


    \[{\left(\frac{\partial G}{\partial P}\right)}_T = V\] and \[{\left(\frac{\partial V}{\partial T}\right)}_P\mathrm{=-}{\left(\frac{\partial S}{\partial P}\right)}_T\]

    • Was this article helpful?