14 added 7 characters in body edited Apr 22 '14 at 19:35 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges where $$p$$ is the total pressure and $$v_i$$ is the molar volume of the $$i^{\text{th}}$$ component in the pure phase. Substituting $$v_{i}$$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $$p$$ as the upper and the partial pressure $$p_{i}$$ as the lower bound for the integration we get $$$$\int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ ,$$$$ so that, introducing the mole fraction $$x_{i}$$, $$$$\mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .$$$$ Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $$a_{i}$$ by scaling the pressure with the (dimensionless) fugacity coefficient $$\varphi_{i}$$ $$$$a_{i} = \frac{\varphi_{i} p_{i}}{p^0}$$$$ where $$p^{0}$$ is the standard pressure for which $$\varphi_{i}=1$$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $$p = p^{0}$$, the following equation arises $$$$\mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .$$$$ Substituting all this togther in our equation for $$\Delta G$$ and noting that the sum of logarithms can be written as a logarithm of products, $$\sum_{i} \ln i = \ln \prod_i i$$, one gets $$$$\Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \log \prod [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod [a_{i}]^{\nu_{i}} \ ,$$$$$$$$\Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,$$$$ where the standard Gibbs free energy of reaction $$\Delta G^{0}$$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $$\Delta G = 0$$ since the system is in equilibrium and then one can introduce the equilibrium constant $$K$$, so that $$$$\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .$$$$ So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $$c_{i}$$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant $$$$K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .$$$$ which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via $$$$K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .$$$$ it is more or less proportional to $$K$$ and thus gives qualitatively the same information. where $$p$$ is the total pressure and $$v_i$$ is the molar volume of the $$i^{\text{th}}$$ component in the pure phase. Substituting $$v_{i}$$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $$p$$ as the upper and the partial pressure $$p_{i}$$ as the lower bound for the integration we get $$$$\int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ ,$$$$ so that, introducing the mole fraction $$x_{i}$$, $$$$\mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .$$$$ Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $$a_{i}$$ by scaling the pressure with the (dimensionless) fugacity coefficient $$\varphi_{i}$$ $$$$a_{i} = \frac{\varphi_{i} p_{i}}{p^0}$$$$ where $$p^{0}$$ is the standard pressure for which $$\varphi_{i}=1$$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $$p = p^{0}$$, the following equation arises $$$$\mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .$$$$ Substituting all this togther in our equation for $$\Delta G$$ and noting that the sum of logarithms can be written as a logarithm of products, $$\sum_{i} \ln i = \ln \prod_i i$$, one gets $$$$\Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \log \prod [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod [a_{i}]^{\nu_{i}} \ ,$$$$ where the standard Gibbs free energy of reaction $$\Delta G^{0}$$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $$\Delta G = 0$$ since the system is in equilibrium and then one can introduce the equilibrium constant $$K$$, so that $$$$\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .$$$$ So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $$c_{i}$$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant $$$$K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .$$$$ which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via $$$$K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .$$$$ it is more or less proportional to $$K$$ and thus gives qualitatively the same information. where $$p$$ is the total pressure and $$v_i$$ is the molar volume of the $$i^{\text{th}}$$ component in the pure phase. Substituting $$v_{i}$$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $$p$$ as the upper and the partial pressure $$p_{i}$$ as the lower bound for the integration we get $$$$\int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ ,$$$$ so that, introducing the mole fraction $$x_{i}$$, $$$$\mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .$$$$ Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $$a_{i}$$ by scaling the pressure with the (dimensionless) fugacity coefficient $$\varphi_{i}$$ $$$$a_{i} = \frac{\varphi_{i} p_{i}}{p^0}$$$$ where $$p^{0}$$ is the standard pressure for which $$\varphi_{i}=1$$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $$p = p^{0}$$, the following equation arises $$$$\mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .$$$$ Substituting all this togther in our equation for $$\Delta G$$ and noting that the sum of logarithms can be written as a logarithm of products, $$\sum_{i} \ln i = \ln \prod_i i$$, one gets $$$$\Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,$$$$ where the standard Gibbs free energy of reaction $$\Delta G^{0}$$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $$\Delta G = 0$$ since the system is in equilibrium and then one can introduce the equilibrium constant $$K$$, so that $$$$\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .$$$$ So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $$c_{i}$$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant $$$$K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .$$$$ which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via $$$$K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .$$$$ it is more or less proportional to $$K$$ and thus gives qualitatively the same information. 13 deleted 4 characters in body edited Apr 22 '14 at 16:28 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges where $$p$$ is the total pressure and $$v_i$$ is the molar volume of the $$i^{\text{th}}$$ component in the pure phase. Substituting $$v_{i}$$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $$p$$ as the upper and the partial pressure $$p_{i}$$ as the lower bound for the integration we get $$$$\int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ ,$$$$ so that, introducing the mole fraction $$x_{i}$$, $$$$\mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .$$$$ Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $$a_{i}$$ by scaling the pressure with the (dimensionless) fugacity coefficient $$\varphi_{i}$$ $$$$a_{i} = \frac{\varphi_{i} p_{i}}{p^0}$$$$ where $$p^{0}$$ is the standard pressure for which $$\varphi_{i}=1$$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $$p = p^{0}$$, the following equation arises $$$$\mu_{i} (p_{i}, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .$$$$$$$$\mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .$$$$ Substituting all this togther in our equation for $$\Delta G$$ and noting that the sum of logarithms can be written as a logarithm of products, $$\sum_{i} \ln i = \ln \prod_i i$$, one gets $$$$\Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \log \prod [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod [a_{i}]^{\nu_{i}} \ ,$$$$ where the standard Gibbs free energy of reaction $$\Delta G^{0}$$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $$\Delta G = 0$$ since the system is in equilibrium and then one can introduce the equilibrium constant $$K$$, so that $$$$\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .$$$$ So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $$c_{i}$$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant $$$$K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .$$$$ which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via $$$$K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .$$$$ it is more or less proportional to $$K$$ and thus gives qualitatively the same information. where $$p$$ is the total pressure and $$v_i$$ is the molar volume of the $$i^{\text{th}}$$ component in the pure phase. Substituting $$v_{i}$$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $$p$$ as the upper and the partial pressure $$p_{i}$$ as the lower bound for the integration we get $$$$\int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ ,$$$$ so that, introducing the mole fraction $$x_{i}$$, $$$$\mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .$$$$ Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $$a_{i}$$ by scaling the pressure with the (dimensionless) fugacity coefficient $$\varphi_{i}$$ $$$$a_{i} = \frac{\varphi_{i} p_{i}}{p^0}$$$$ where $$p^{0}$$ is the standard pressure for which $$\varphi_{i}=1$$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $$p = p^{0}$$, the following equation arises $$$$\mu_{i} (p_{i}, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .$$$$ Substituting all this togther in our equation for $$\Delta G$$ and noting that the sum of logarithms can be written as a logarithm of products, $$\sum_{i} \ln i = \ln \prod_i i$$, one gets $$$$\Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \log \prod [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod [a_{i}]^{\nu_{i}} \ ,$$$$ where the standard Gibbs free energy of reaction $$\Delta G^{0}$$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $$\Delta G = 0$$ since the system is in equilibrium and then one can introduce the equilibrium constant $$K$$, so that $$$$\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .$$$$ So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $$c_{i}$$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant $$$$K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .$$$$ which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via $$$$K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .$$$$ it is more or less proportional to $$K$$ and thus gives qualitatively the same information. where $$p$$ is the total pressure and $$v_i$$ is the molar volume of the $$i^{\text{th}}$$ component in the pure phase. Substituting $$v_{i}$$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $$p$$ as the upper and the partial pressure $$p_{i}$$ as the lower bound for the integration we get $$$$\int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ ,$$$$ so that, introducing the mole fraction $$x_{i}$$, $$$$\mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .$$$$ Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $$a_{i}$$ by scaling the pressure with the (dimensionless) fugacity coefficient $$\varphi_{i}$$ $$$$a_{i} = \frac{\varphi_{i} p_{i}}{p^0}$$$$ where $$p^{0}$$ is the standard pressure for which $$\varphi_{i}=1$$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $$p = p^{0}$$, the following equation arises $$$$\mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .$$$$ Substituting all this togther in our equation for $$\Delta G$$ and noting that the sum of logarithms can be written as a logarithm of products, $$\sum_{i} \ln i = \ln \prod_i i$$, one gets $$$$\Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \log \prod [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod [a_{i}]^{\nu_{i}} \ ,$$$$ where the standard Gibbs free energy of reaction $$\Delta G^{0}$$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $$\Delta G = 0$$ since the system is in equilibrium and then one can introduce the equilibrium constant $$K$$, so that $$$$\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .$$$$ So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $$c_{i}$$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant $$$$K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .$$$$ which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via $$$$K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .$$$$ it is more or less proportional to $$K$$ and thus gives qualitatively the same information. 12 added 1 character in body edited Apr 22 '14 at 11:29 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges but since $$\mu^{*}_{i}$$ is associated with a pure facephase, $$\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$$ can be simplified to $$$$\left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i}$$$$ but since $$\mu^{*}_{i}$$ is associated with a pure face, $$\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$$ can be simplified to $$$$\left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i}$$$$ but since $$\mu^{*}_{i}$$ is associated with a pure phase, $$\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$$ can be simplified to $$$$\left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i}$$$$ 11 deleted 2 characters in body edited Apr 22 '14 at 10:59 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges 10 edited body edited Apr 22 '14 at 9:19 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges 9 changed log to ln in order to avoid confusion with the decadic logarithm edited Apr 22 '14 at 9:06 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges 8 added details for the Maxwell relation used, corrected spelling edited Feb 19 '14 at 23:12 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges 7 added 24 characters in body edited Aug 8 '13 at 14:44 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges 6 added 791 characters in body edited Aug 8 '13 at 14:28 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges 5 added 791 characters in body edited Aug 8 '13 at 14:22 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges 4 added 147 characters in body edited Aug 7 '13 at 21:14 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges 3 added 105 characters in body edited Aug 7 '13 at 18:36 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges 2 added 295 characters in body edited Aug 7 '13 at 16:52 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges 1 answered Aug 7 '13 at 16:46 Philipp 15.1k22 gold badges5858 silver badges105105 bronze badges