Question
Check by a direct substitution that the equation (4) at $c=c_0, \kappa=\kappa_0 T^\sigma$ is an invariant relative transformation of expansion-compression under the condition of $\alpha \beta^{-2} \gamma^\sigma=1$.
Step 1
We have an equation (4) with parameters $c=c_0$ and $\kappa=\kappa_0 T^\sigma$. We need to verify that this equation is invariant under expansion-compression transformations when the condition $\alpha\beta^{-2}\gamma^\sigma=1$ is satisfied. Show more…
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$\begin{aligned} \dot{Q}=\kappa \frac{\partial T}{\partial x}=-A \sqrt{T} \frac{\partial T}{\partial x} \\=-\frac{2}{3} A \frac{\partial T^{3 / 2}}{\partial x},(A=\text { constant }) \\ &=\frac{2}{3} A \frac{\left(T_{1}^{3 / 2}-T_{2}^{3 / 2}\right)}{l} \\ \text { Thus } \\ \text { or using } \\ T^{3 / 2}=T_{1}^{3 / 2}+\frac{x}{l}\left(T_{2}^{3 / 2}-T_{1}^{3 / 2}\right) \text { or }\left(\frac{T}{T_{1}}\right)^{3 / 2}=1+\frac{x}{l}\left(\left(\frac{T_{2}}{T}\right)^{3 / 2}\right.&-1) \\ T=T_{1}\left[1+\frac{x}{l}\left\{\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2}\right.\right.&-1\}] \end{aligned}$
Thermodynamics And Molecular Physics
Transport Phenomena
A substance for which $\mathrm{k}$ is a constant undergoes an isothermal, mechanically reversible process from initial state $\left(P_{1}, V_{1}\right)$ to final state $\left(P_{2}, V_{2}\right),$ where $V$ is molar volume. (a) Starting with the definition of $\kappa$, show that the path of the process is described by: $$V=A(T) \exp (-\kappa P)$$ (b) Determine an exact expression which gives the isothermal work done on 1 mol of this constant-K substance.
Because of the linearity of the section $B C$ whose equation is $\frac{p}{p_{0}}=\frac{v V}{V_{0}}(=p=\alpha V)$ We have $\frac{\tau}{v}=v$ or $v=\sqrt{\tau}$ Here $\quad Q_{2}^{\prime \prime}=C_{V} T_{0}(\sqrt{\tau}-1)$ $Q_{2}^{\prime \prime \prime}=C_{p} T_{0}\left(1-\frac{1}{\sqrt{\tau}}\right)=C_{p} \frac{T_{0}}{\sqrt{\tau}}(\sqrt{\tau}-1)$ Thus $Q_{2}^{\prime}=Q_{2}^{\prime \prime}+Q_{2}^{\prime \prime \prime}=\frac{R T_{0}}{\gamma-1}(\sqrt{\tau}-1)\left(1+\frac{\gamma}{\sqrt{\tau}}\right)$ Along $B C$, the specific heat $C$ is given by $C d T=C_{V} d T+p d V=C_{V} d T+d\left(\frac{1}{2} \alpha V^{2}\right)=\left(C_{V}+\frac{1}{2} R\right) d T$ Thus $Q_{1}=\frac{1}{2} R T_{0} \frac{\gamma+1}{\gamma-1} \frac{\tau-1}{\sqrt{\tau}}$ Finally $\eta=1-\frac{Q_{2}^{\prime}}{Q_{1}}=1-2 \frac{\sqrt{\tau}+\gamma}{\sqrt{\tau}+1} \frac{1}{\gamma+1}=\frac{(\gamma-1)(\sqrt{\tau}-1)}{(\gamma+1)(\sqrt{\tau}+1)}$
The Second Law of Thermodynamics. Entropy
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