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 2 reference added edited Jun 17 '18 at 16:37 porphyrin 20k11 gold badge3535 silver badges6060 bronze badges (a) The idea of potentials is familiar from mechanical and electrical systems. In an electric field the work required to move a charge $$q$$ from one location with potential $$\theta_1$$ to one with $$\theta_2$$ is $$q(\theta_2-\theta_1)$$. This expression has the form of a constant factor ($$q$$) times the change in potential. In a gravitational field the potential at height $$h$$ is $$gh$$ and so the work needed to lift a mass $$m$$ (the constant factor) from $$h_1 \to h_2$$ is $$mg(h_2-h_1)= mg\Delta h$$. We also know that these systems naturally move to position of lowest potential energy to reach equilibrium: stones do drop to the ground. In a chemical system we can choose the mole number $$n$$ as the constant factor and then the molar free energy is the chemical potential which is usually given the symbol $$\mu$$. The chemical potential controls mass equilibrium just as temperature controls thermal equilibrium. If there is a gradient of chemical potential then mass flows (or molecules rearrange). At equilibrium the chemical potential of all parts are equal. Similarly energy flows in a temperature gradient until equilibrium is achieved and the temperature is uniform throughout. In this sense the chemical potential is considered as providing the 'force' to drive chemical systems to equilibrium. (b) If we wish to transfer $$n$$ moles of a gas from a state with molar free energy $$G_1$$ to one of $$G_2$$ then the work done (other than expansion) is the change in free energy $$n(G_2-G_1)$$. Suppose now that we isothermally transfer $$n$$ moles of an ideal gas from a vessel with pressure $$p_1$$ to one with pressure $$p_2$$ the work involved is $$nRT\ln(p_2/p_1)$$ which is also $$n(G_2-G_1)$$ so that $$(G_2-G_1)=RT\ln(p_2/p_1)$$. Usually we relate the pressures to one at a standard state, say 1 atm. then letting $$G_1 \to G^\mathrm{o}$$ and $$G_2 \to G$$ then $$G= G^\mathrm{o}+RT\ln(p)$$ where $$p$$ is $$p/(1\text{atm})$$ which is a dimensionless ratio. As the molar free energy is the chemical potential, $$\mu=\mu^{\mathrm{o}}+RT\ln(p)$$ (The equation $$(G_2-G_1)=RT\ln(p_2/p_1)$$ can also be derived by starting with $$dG=Vdp-SdT$$, at constant temperature ($$dT=0$$) and for an ideal gas by using the gas law substitute for $$V$$, and integrating from $$p_1\to p_2$$.) (c) More formally the chemical potential can be defined via the work done in reversibly compressing a gas composed of $$i$$ species with mole numbers $$n_i$$. If the force $$F$$ is applied to a piston that moves a distance $$dx$$ then $$Fdx= -pdV +\sum_i \mu_i dn_i$$ where $$\mu_i$$ is the chemical potential of species $$i$$ and is defined by this equation. This means that the chemical potential is the (reversible) rate of change of internal energy with mole number while keeping other variables ($$S,V$$) constant, thus since $$dU=TdS-pdV+\sum_i \mu_i dn_i$$, where $$U$$ is the internal energy, then $$\displaystyle \mu_i= \left(\frac{\partial U}{\partial n_i}\right)_{T,S,n_j}$$. (The subscript $$n_j$$ means that other mole number are held constant.) It is more common, however, do define the chemical potential in terms of the Gibbs free energy and where $$T,p$$ are held constant. For a pure substance your textbook will derive the equation $$dG=Vdp-SdT$$ but if there is a mixture the number of moles can vary then it is necessary to account for energy changes due to this by adding a term to the energy and doing this produces $$dG=Vdp-SdT+\sum_i \mu_i dn_i$$ which is sometimes referred to as the fundamental equation of chemical thermodynamics, and then the chemical potential is defined as $$\displaystyle \mu_i= \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_j}$$ [discussion in (a) and (b) based on arguments in Lewis & Randall, 'Thermodynamics' publ McGraw-Hill.] (a) The idea of potentials is familiar from mechanical and electrical systems. In an electric field the work required to move a charge $$q$$ from one location with potential $$\theta_1$$ to one with $$\theta_2$$ is $$q(\theta_2-\theta_1)$$. This expression has the form of a constant factor ($$q$$) times the change in potential. In a gravitational field the potential at height $$h$$ is $$gh$$ and so the work needed to lift a mass $$m$$ (the constant factor) from $$h_1 \to h_2$$ is $$mg(h_2-h_1)= mg\Delta h$$. We also know that these systems naturally move to position of lowest potential energy to reach equilibrium: stones do drop to the ground. In a chemical system we can choose the mole number $$n$$ as the constant factor and then the molar free energy is the chemical potential which is usually given the symbol $$\mu$$. The chemical potential controls mass equilibrium just as temperature controls thermal equilibrium. If there is a gradient of chemical potential then mass flows (or molecules rearrange). At equilibrium the chemical potential of all parts are equal. Similarly energy flows in a temperature gradient until equilibrium is achieved and the temperature is uniform throughout. In this sense the chemical potential is considered as providing the 'force' to drive chemical systems to equilibrium. (b) If we wish to transfer $$n$$ moles of a gas from a state with molar free energy $$G_1$$ to one of $$G_2$$ then the work done (other than expansion) is the change in free energy $$n(G_2-G_1)$$. Suppose now that we isothermally transfer $$n$$ moles of an ideal gas from a vessel with pressure $$p_1$$ to one with pressure $$p_2$$ the work involved is $$nRT\ln(p_2/p_1)$$ which is also $$n(G_2-G_1)$$ so that $$(G_2-G_1)=RT\ln(p_2/p_1)$$. Usually we relate the pressures to one at a standard state, say 1 atm. then letting $$G_1 \to G^\mathrm{o}$$ and $$G_2 \to G$$ then $$G= G^\mathrm{o}+RT\ln(p)$$ where $$p$$ is $$p/(1\text{atm})$$ which is a dimensionless ratio. As the molar free energy is the chemical potential, $$\mu=\mu^{\mathrm{o}}+RT\ln(p)$$ (The equation $$(G_2-G_1)=RT\ln(p_2/p_1)$$ can also be derived by starting with $$dG=Vdp-SdT$$, at constant temperature ($$dT=0$$) and for an ideal gas by using the gas law substitute for $$V$$, and integrating from $$p_1\to p_2$$.) (c) More formally the chemical potential can be defined via the work done in reversibly compressing a gas composed of $$i$$ species with mole numbers $$n_i$$. If the force $$F$$ is applied to a piston that moves a distance $$dx$$ then $$Fdx= -pdV +\sum_i \mu_i dn_i$$ where $$\mu_i$$ is the chemical potential of species $$i$$ and is defined by this equation. This means that the chemical potential is the (reversible) rate of change of internal energy with mole number while keeping other variables ($$S,V$$) constant, thus since $$dU=TdS-pdV+\sum_i \mu_i dn_i$$, where $$U$$ is the internal energy, then $$\displaystyle \mu_i= \left(\frac{\partial U}{\partial n_i}\right)_{T,S,n_j}$$. (The subscript $$n_j$$ means that other mole number are held constant.) It is more common, however, do define the chemical potential in terms of the Gibbs free energy and where $$T,p$$ are held constant. For a pure substance your textbook will derive the equation $$dG=Vdp-SdT$$ but if there is a mixture the number of moles can vary then it is necessary to account for energy changes due to this by adding a term to the energy and doing this produces $$dG=Vdp-SdT+\sum_i \mu_i dn_i$$ which is sometimes referred to as the fundamental equation of chemical thermodynamics, and then the chemical potential is defined as $$\displaystyle \mu_i= \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_j}$$ (a) The idea of potentials is familiar from mechanical and electrical systems. In an electric field the work required to move a charge $$q$$ from one location with potential $$\theta_1$$ to one with $$\theta_2$$ is $$q(\theta_2-\theta_1)$$. This expression has the form of a constant factor ($$q$$) times the change in potential. In a gravitational field the potential at height $$h$$ is $$gh$$ and so the work needed to lift a mass $$m$$ (the constant factor) from $$h_1 \to h_2$$ is $$mg(h_2-h_1)= mg\Delta h$$. We also know that these systems naturally move to position of lowest potential energy to reach equilibrium: stones do drop to the ground. In a chemical system we can choose the mole number $$n$$ as the constant factor and then the molar free energy is the chemical potential which is usually given the symbol $$\mu$$. The chemical potential controls mass equilibrium just as temperature controls thermal equilibrium. If there is a gradient of chemical potential then mass flows (or molecules rearrange). At equilibrium the chemical potential of all parts are equal. Similarly energy flows in a temperature gradient until equilibrium is achieved and the temperature is uniform throughout. In this sense the chemical potential is considered as providing the 'force' to drive chemical systems to equilibrium. (b) If we wish to transfer $$n$$ moles of a gas from a state with molar free energy $$G_1$$ to one of $$G_2$$ then the work done (other than expansion) is the change in free energy $$n(G_2-G_1)$$. Suppose now that we isothermally transfer $$n$$ moles of an ideal gas from a vessel with pressure $$p_1$$ to one with pressure $$p_2$$ the work involved is $$nRT\ln(p_2/p_1)$$ which is also $$n(G_2-G_1)$$ so that $$(G_2-G_1)=RT\ln(p_2/p_1)$$. Usually we relate the pressures to one at a standard state, say 1 atm. then letting $$G_1 \to G^\mathrm{o}$$ and $$G_2 \to G$$ then $$G= G^\mathrm{o}+RT\ln(p)$$ where $$p$$ is $$p/(1\text{atm})$$ which is a dimensionless ratio. As the molar free energy is the chemical potential, $$\mu=\mu^{\mathrm{o}}+RT\ln(p)$$ (The equation $$(G_2-G_1)=RT\ln(p_2/p_1)$$ can also be derived by starting with $$dG=Vdp-SdT$$, at constant temperature ($$dT=0$$) and for an ideal gas by using the gas law substitute for $$V$$, and integrating from $$p_1\to p_2$$.) (c) More formally the chemical potential can be defined via the work done in reversibly compressing a gas composed of $$i$$ species with mole numbers $$n_i$$. If the force $$F$$ is applied to a piston that moves a distance $$dx$$ then $$Fdx= -pdV +\sum_i \mu_i dn_i$$ where $$\mu_i$$ is the chemical potential of species $$i$$ and is defined by this equation. This means that the chemical potential is the (reversible) rate of change of internal energy with mole number while keeping other variables ($$S,V$$) constant, thus since $$dU=TdS-pdV+\sum_i \mu_i dn_i$$, where $$U$$ is the internal energy, then $$\displaystyle \mu_i= \left(\frac{\partial U}{\partial n_i}\right)_{T,S,n_j}$$. (The subscript $$n_j$$ means that other mole number are held constant.) It is more common, however, do define the chemical potential in terms of the Gibbs free energy and where $$T,p$$ are held constant. For a pure substance your textbook will derive the equation $$dG=Vdp-SdT$$ but if there is a mixture the number of moles can vary then it is necessary to account for energy changes due to this by adding a term to the energy and doing this produces $$dG=Vdp-SdT+\sum_i \mu_i dn_i$$ which is sometimes referred to as the fundamental equation of chemical thermodynamics, and then the chemical potential is defined as $$\displaystyle \mu_i= \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_j}$$ [discussion in (a) and (b) based on arguments in Lewis & Randall, 'Thermodynamics' publ McGraw-Hill.] 1 answered Jun 17 '18 at 16:20 porphyrin 20k11 gold badge3535 silver badges6060 bronze badges (a) The idea of potentials is familiar from mechanical and electrical systems. In an electric field the work required to move a charge $$q$$ from one location with potential $$\theta_1$$ to one with $$\theta_2$$ is $$q(\theta_2-\theta_1)$$. This expression has the form of a constant factor ($$q$$) times the change in potential. In a gravitational field the potential at height $$h$$ is $$gh$$ and so the work needed to lift a mass $$m$$ (the constant factor) from $$h_1 \to h_2$$ is $$mg(h_2-h_1)= mg\Delta h$$. We also know that these systems naturally move to position of lowest potential energy to reach equilibrium: stones do drop to the ground. In a chemical system we can choose the mole number $$n$$ as the constant factor and then the molar free energy is the chemical potential which is usually given the symbol $$\mu$$. The chemical potential controls mass equilibrium just as temperature controls thermal equilibrium. If there is a gradient of chemical potential then mass flows (or molecules rearrange). At equilibrium the chemical potential of all parts are equal. Similarly energy flows in a temperature gradient until equilibrium is achieved and the temperature is uniform throughout. In this sense the chemical potential is considered as providing the 'force' to drive chemical systems to equilibrium. (b) If we wish to transfer $$n$$ moles of a gas from a state with molar free energy $$G_1$$ to one of $$G_2$$ then the work done (other than expansion) is the change in free energy $$n(G_2-G_1)$$. Suppose now that we isothermally transfer $$n$$ moles of an ideal gas from a vessel with pressure $$p_1$$ to one with pressure $$p_2$$ the work involved is $$nRT\ln(p_2/p_1)$$ which is also $$n(G_2-G_1)$$ so that $$(G_2-G_1)=RT\ln(p_2/p_1)$$. Usually we relate the pressures to one at a standard state, say 1 atm. then letting $$G_1 \to G^\mathrm{o}$$ and $$G_2 \to G$$ then $$G= G^\mathrm{o}+RT\ln(p)$$ where $$p$$ is $$p/(1\text{atm})$$ which is a dimensionless ratio. As the molar free energy is the chemical potential, $$\mu=\mu^{\mathrm{o}}+RT\ln(p)$$ (The equation $$(G_2-G_1)=RT\ln(p_2/p_1)$$ can also be derived by starting with $$dG=Vdp-SdT$$, at constant temperature ($$dT=0$$) and for an ideal gas by using the gas law substitute for $$V$$, and integrating from $$p_1\to p_2$$.) (c) More formally the chemical potential can be defined via the work done in reversibly compressing a gas composed of $$i$$ species with mole numbers $$n_i$$. If the force $$F$$ is applied to a piston that moves a distance $$dx$$ then $$Fdx= -pdV +\sum_i \mu_i dn_i$$ where $$\mu_i$$ is the chemical potential of species $$i$$ and is defined by this equation. This means that the chemical potential is the (reversible) rate of change of internal energy with mole number while keeping other variables ($$S,V$$) constant, thus since $$dU=TdS-pdV+\sum_i \mu_i dn_i$$, where $$U$$ is the internal energy, then $$\displaystyle \mu_i= \left(\frac{\partial U}{\partial n_i}\right)_{T,S,n_j}$$. (The subscript $$n_j$$ means that other mole number are held constant.) It is more common, however, do define the chemical potential in terms of the Gibbs free energy and where $$T,p$$ are held constant. For a pure substance your textbook will derive the equation $$dG=Vdp-SdT$$ but if there is a mixture the number of moles can vary then it is necessary to account for energy changes due to this by adding a term to the energy and doing this produces $$dG=Vdp-SdT+\sum_i \mu_i dn_i$$ which is sometimes referred to as the fundamental equation of chemical thermodynamics, and then the chemical potential is defined as $$\displaystyle \mu_i= \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_j}$$