Because \(G\) is a state function, the change in \(G\) depends only on the initial and final states, not on the path taken between them. This means we are free to choose any convenient pathway for calculating \(\Delta G\).
To see how this works, begin with the definition of the Gibbs function:
\[ G = U+pV-TS \]
Differentiating gives
\[ dG = dU + p\,dV + V\,dp - T\,dS - S\,dT \]
Now substitute the combined First and Second Law:
\[ dU = T\,dS - p\,dV \]
This gives
\[ dG = (T\,dS-p\,dV) + p\,dV + V\,dp - T\,dS - S\,dT \]
\[ \boxed{ dG = V\,dp - S\,dT } \]
This expression shows that the natural variables of \(G\) are \(p\) and \(T\):
\[ G = G(p,T) \]
Since \(G=G(p,T)\), its total differential can also be written as
\[ dG = \left( \frac{\partial G}{\partial p} \right)_T dp + \left( \frac{\partial G}{\partial T} \right)_p dT \]
Comparing this expression with
\[ dG = V\,dp - S\,dT \]
gives
\[ \boxed{ \left( \frac{\partial G}{\partial p} \right)_T = V } \]
\[ \boxed{ \left( \frac{\partial G}{\partial T} \right)_p = -S } \]
These two derivatives tell us how \(G\) changes when pressure changes at constant temperature and when temperature changes at constant pressure.
Suppose a system changes from an initial state \((p_1,T_1)\) to a final state \((p_2,T_2)\).
Instead of trying to change both variables at once, we can choose a two-step pathway:
Since \(G\) is a state function, this pathway gives the same \(\Delta G\) as any other pathway connecting the same initial and final states.
At constant temperature,
\[ dT = 0 \]
so
\[ dG = V\,dp \]
Integrating from \(p_1\) to \(p_2\):
\[ \Delta G_{\mathrm{pressure}} = \int_{p_1}^{p_2} V\,dp \]
At constant pressure,
\[ dp = 0 \]
so
\[ dG = -S\,dT \]
Integrating from \(T_1\) to \(T_2\):
\[ \Delta G_{\mathrm{temperature}} = -\int_{T_1}^{T_2} S\,dT \]
The total change in \(G\) is the sum of the two contributions:
\[ \Delta G = \Delta G_{\mathrm{pressure}} + \Delta G_{\mathrm{temperature}} \]
Therefore,
\[ \boxed{ \Delta G = \int_{p_1}^{p_2} V\,dp - \int_{T_1}^{T_2} S\,dT } \]
This expression is useful because it allows a simultaneous change in pressure and temperature to be broken into two simpler steps.
Big picture: Because \(G\) is a state function, we do not need to follow the actual path taken by the system. We can choose a convenient pathway, calculate the pressure and temperature contributions separately, and add them to obtain the total \(\Delta G\).
Calculate \(\Delta G\) for \(1.50\ \mathrm{mol}\) of an ideal gas \(\left(C_p=\frac{5}{2}R\right)\) changing from \(p_1=1.00\ \mathrm{atm}\), \(T_1=300\ \mathrm{K}\) to \(p_2=0.333\ \mathrm{atm}\), \(T_2=450\ \mathrm{K}\).
Assume the molar entropy at the initial state is
\[ S_{m,1}=155\ \mathrm{J\,mol^{-1}\,K^{-1}} \]
Since \(G=G(p,T)\), use
\[ dG=Vdp-SdT \]
For an ideal gas,
\[ V=\frac{nRT}{p} \]
so
\[ \Delta G_p = \int_{p_1}^{p_2}Vdp = nRT_1\ln\left(\frac{p_2}{p_1}\right) \]
\[ \Delta G_p = (1.50)(8.314)(300) \ln\left(\frac{0.333}{1.00}\right) \]
\[ \Delta G_p = -4.12\times10^3\ \mathrm{J} \]
First calculate the entropy after the isothermal pressure change:
\[ S_B = S_1 - nR\ln\left(\frac{p_2}{p_1}\right) \]
\[ S_B = (1.50)(155) - (1.50)(8.314)\ln(0.333) \]
\[ S_B = 246\ \mathrm{J\,K^{-1}} \]
At constant pressure,
\[ S(T) = S_B+nC_p\ln\left(\frac{T}{T_1}\right) \]
Therefore,
\[ \Delta G_T = -\int_{T_1}^{T_2}S(T)dT \]
\[ \Delta G_T = -S_B(T_2-T_1) - nC_p \left[ T_2\ln\left(\frac{T_2}{T_1}\right) - T_2+T_1 \right] \]
\[ \Delta G_T = -(246)(150) - (1.50)\left(\frac{5}{2}R\right) \left[ 450\ln\left(\frac{450}{300}\right)-450+300 \right] \]
\[ \Delta G_T = -3.79\times10^4\ \mathrm{J} \]
\[ \Delta G = \Delta G_p+\Delta G_T \]
\[ \Delta G = -4.12\ \mathrm{kJ} - 37.9\ \mathrm{kJ} \]
\[ \boxed{ \Delta G = -42.0\ \mathrm{kJ} } \]
Interpretation: The pressure decrease lowers \(G\) for an ideal gas, and the temperature increase also lowers \(G\) because \(\left(\partial G/\partial T\right)_p=-S\). The total change is therefore negative.
Calculate the change in internal energy for \(1.50\ \mathrm{mol}\) of a monatomic ideal gas \(\left(C_V=\frac{3}{2}R\right)\) as it changes from \(p_1=1.00\ \mathrm{atm}\), \(T_1=300\ \mathrm{K}\) to \(p_2=0.333\ \mathrm{atm}\), \(T_2=450\ \mathrm{K}\).
For an ideal gas,
\[ \left( \frac{\partial U} {\partial V} \right)_T = 0 \]
Therefore, the internal energy depends only on temperature.
The total differential of the internal energy is
\[ dU = \left( \frac{\partial U} {\partial T} \right)_V dT \]
and since
\[ \left( \frac{\partial U} {\partial T} \right)_V = nC_V \]
it follows that
\[ \Delta U = nC_V\Delta T \]
For a monatomic ideal gas,
\[ C_V = \frac{3}{2}R = \frac{3}{2}(8.314) = 12.47\ \mathrm{J\,mol^{-1}\,K^{-1}} \]
The temperature change is
\[ \Delta T = 450\ \mathrm{K} - 300\ \mathrm{K} = 150\ \mathrm{K} \]
Therefore,
\[ \Delta U = (1.50\ \mathrm{mol}) (12.47\ \mathrm{J\,mol^{-1}\,K^{-1}}) (150\ \mathrm{K}) \]
\[ \Delta U = 2.81\times10^3\ \mathrm{J} \]
\[ \boxed{ \Delta U = +2.81\ \mathrm{kJ} } \]
Interpretation: Although both the pressure and volume change during the process, neither quantity appears in the final calculation. For an ideal gas, the internal energy depends only on temperature, so the entire change in internal energy is determined by the \(150\ \mathrm{K}\) temperature increase.
Big picture: Because thermodynamic functions are state functions, we are free to choose convenient pathways when calculating changes. By decomposing a simultaneous change in pressure and temperature into separate pressure and temperature contributions, complex calculations become much easier while still producing the correct overall change in Gibbs free energy.