这是一份nottingham诺丁汉大学PHYS2002作业代写的成功案例

$$
G\left(n_{1}, n_{2}, \ldots n_{N}\right)=\sum_{i=1}^{N} \Delta_{1} G_{\mathrm{m}, i}+R T \sum_{i=1}^{N} n_{i} \ln \left(n_{i} / \sum_{j=1}^{N} n_{j}\right)
$$
where $N$ is the numberr of components, $n_{i}$ is the molar amount of $i$-th component, $T$ is the system temperature (i.e. the glass transition temperature, $T_{\mathrm{g}}$, for particular glass) and $\Delta_{\mathrm{f}} G_{\mathrm{m}, i}$ is the molar Gibbs formation energy of pure $i$-th component at the pressure of the system and temperature $T$. The system components are ordered such way that $X_{i}(i=1,2, \ldots M<N)$ are pure oxides and $X_{i}(i=M+1, M+2, \ldots N)$ are compounds formed from oxides by reversible reactions
$$
X_{i} \leftrightarrow \sum_{j=1}^{M} v_{i, j} X_{j}, \quad i=M+1, M+2, \ldots N
$$
Let us suppose the system composition given by the molar amounts of pure unreacted oxides $n_{0, i}(i=1,2, \ldots . M)$. Then the mass balance constraints can be written in the form:
$$
n_{0, j}=n_{j}+\sum_{i=M+1}^{N} v_{i, j} n_{i}, \quad j=1,2, \ldots M
$$

PHYS2002 COURSE NOTES ：

$$
\lg \eta=\lg \eta_{o}+\frac{1}{2.3}\left(\frac{H^{}-H(T)}{R T}-\frac{S^{}-S(T)}{R}\right)
$$
Here $\mathrm{H}^{}$ and $\mathrm{S}^{}$ are the enthalpy and the entropy of the active state. Under the assumption that non-equilibrium viscosity corresponds to a system with structure fixed at $\mathrm{T}=\mathrm{T}{\mathrm{f}}$, i.e. with constant enthalpy $\mathrm{H}{\mathrm{f}}$ and constant entropy $\mathrm{S}{\mathrm{f}}$, the difference $\mathrm{m}-\mathrm{m}{\mathrm{g}}$ becomes:
$$
m-m_{f}=\frac{1}{2.3 R}\left(\left.\frac{1}{T} \frac{d H(T)}{d\left(T_{f} / T\right)}\right|{T=T{f}}-\left.\frac{d S(T)}{d\left(T_{f} / T\right)}\right|{T=T{f}}\right)_{\text {equilibrium }}
$$