Added Luis's English solution of 5.9.10

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+ <title>Savchenko Solutions</title>
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+ <meta property="og:title" content="Savchenko Solutions"/>
+ <meta name="description"
+ content="A website with solutions to physics problems from Savenko Textbook" />
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+ <header>
+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <main>
+ <article>
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h4>Statement</h4>
+ <p>
+ $5.9.10^*$
+ Find the efficience of cycles shown in following figures, if as power agent it is taken a monoatomic ideal gas.
+ </p>
+ <center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" alt="5.9.10" width="350" />
+ <figcaption>
+ For problem 5.9.10
+ </figcaption>
+ </figure>
+ </center>
+ <h4>Solution</h4>
+ <p class="TxtSolutions">
+ a) Let enumerate the states from 1 to 4 in clockwise sense starting in the coordinate $(P_1,V_1)$. The efficience is defined as
+ </p>
+ <center>
+ $\eta = 1-\frac{|Q_c|}{Q_h}=1-\frac{|Q_{34}|}{Q_{12}} \;(1)$
+ </center>
+ <p class="TxtSolutions">
+ where $Q_{34}$ is the heat released by gas (cold reservoir) and $Q_{12}$ is the heat absorbed by gas (hot reservoir), while $Q_{23}$ and $Q_{41}$ are zero (adiabatic processes). Since $3-4$ is an isochoric process,
+ </p>
+ <center>
+ $Q_{34}=\Delta U_{34} = nC_v\left(T_4-T_3\right) \;(2)$
+ </center>
+ <p class="TxtSolutions">
+ From Poisson Law
+ </p>
+ <center>
+ $P_1V_1^\gamma = P_4V_2^\gamma$
+ </center>
+ <p class="TxtSolutions">
+ and the state 4 is characterized by
+ </p>
+ <center>
+ $P_4V_2=nRT_4$
+ </center>
+ <p class="TxtSolutions">
+ So, separating $P_4$ and putting it into above Poisson law
+ </p>
+ <center>
+ $T_4 = \frac{P_1V_1^\gamma}{nRV_2^{\gamma-1}} \;(3)$
+ </center>
+ <p class="TxtSolutions">
+ Proceeding similarly, $T_3$ can be found. Applying Poisson Law for process $2-3$,
+ </p>
+ <center>
+ $P_2V_1^\gamma=P_3V_2^\gamma$
+ </center>
+ <p class="TxtSolutions">
+ State equation for state 3 is
+ </p>
+ <center>
+ $P_3V_2=nRT_3$
+ </center>
+ <p class="TxtSolutions">
+ Separating $P_3$ and substituting into respective Poisson Law
+ </p>
+ <center>
+ $T_3 = \frac{P_2V_1^\gamma}{nRV_2^{\gamma-1}} \;(4)$
+ </center>
+ <p class="TxtSolutions">
+ Substituting $(3)$ and $(4)$ into $(2)$, and as $C_v=iR/2$
+ </p>
+ <center>
+ $Q_{34} = \frac{i}{2}\frac{V_1^\gamma}{V_2^{\gamma-1}}\left(P_1-P_2\right) \;(5)$
+ </center>
+ <p class="TxtSolutions">
+ For the process $1-2$
+ </p>
+ <center>
+ $Q_{12}=\Delta U_{12} = nC_v\left(T_2-T_1\right)$
+ </center>
+ <p class="TxtSolutions">
+ Taking account the state equations for states 1 and 2, temperatures can be found,
+ </p>
+ <center>
+ $Q_{12}=nC_v\left(\frac{P_2V_1}{nR}-\frac{P_1V_1}{nR}\right)$
+ </center>
+ <center>
+ $Q_{12}=\frac{i}{2}V_1\left(P_2-P_1\right) \;(6)$
+ </center>
+ <p class="TxtSolutions">
+ Putting $(5)$ and $(6)$ into $(1)$
+ </p>
+ <center>
+ $\eta = 1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}$
+ </center>
+ <p class="TxtSolutions">
+ Since $\gamma=C_p/C_v=1+2/i$ and $i=3$ (monoatomic gas),
+ </p>
+ <center>
+ $\boxed{\eta = 1-\left(\frac{V_1}{V_2}\right)^{\frac{2}{3}}}$
+ </center>
+ <p class="TxtSolutions">
+ b) Let enumerate again the states following a clockwise way, this time starting in coordinate $(P_1,T_1)$. The efficience is
+ </p>
+ <center>
+ $\eta = 1-\frac{|Q_c|}{Q_h} \;(I)$
+ </center>
+ <p class="TxtSolutions">
+ For identifying correctly the reservoirs, let's see the following graph corresponding to the cycle in a $P$ vs $V$ plot.
+ </p>
+ <center>
+ <figure>
+ <img src="PVb.png"
+ loading="lazy" alt="5.9.10" width="300" />
+ <figcaption>
+ $P$ vs $V$ plot
+ </figcaption>
+ </figure>
+ </center>
+ <p class="TxtSolutions">
+ The hot reservoir is defined as
+ </p>
+ <center>
+ $Q_h = Q_{23} + Q_{34} \;(II)$
+ </center>
+ <p class="TxtSolutions">
+ and cold one
+ </p>
+ <center>
+ $Q_c = Q_{12} + Q_{41} \;(III)$
+ </center>
+ <p class="TxtSolutions">
+ Process $2-3$ is isobaric, then
+ </p>
+ <center>
+ $Q_{23}=nC_p\left(T_2-T_1\right) \;(IV)$
+ </center>
+ <p class="TxtSolutions">
+ Process $3-4$ is isothermal, so
+ </p>
+ <center>
+ $Q_{34} = A_{34} = nRT_2\ln{\frac{V_4}{V_3}}$
+ </center>
+ <p class="TxtSolutions">
+ Since $V=nRT/P$ according state equation
+ </p>
+ <center>
+ $Q_{34} = nRT_2\ln{\frac{P_2}{P_1}} \;(V)$
+ </center>
+ <p class="TxtSolutions">
+ Substituting $(IV)$ and $(V)$ into $(II)$and taking account that $C_p=(i+2)R/2$
+ </p>
+ <center>
+ $Q_h = \frac{i+2}{2}nR\left(T_2-T_1\right)+nRT_2\ln{\frac{P_2}{P_1}} \;(VI)$
+ </center>
+ <p class="TxtSolutions">
+ Process $1-2$ is isothermal, so
+ </p>
+ <center>
+ $Q_{12}=A_{12}=nRT_1\ln{\frac{V_2}{V_1}}$
+ </center>
+ <p class="TxtSolutions">
+ and newly, taking account state equation
+ </p>
+ <center>
+ $Q_{12}=A_{12}=nRT_1\ln{\frac{P_1}{P_2} \;(VII)}$
+ </center>
+ <p class="TxtSolutions">
+ Finally, process $4-1$ is isobaric
+ </p>
+ <center>
+ $Q_{14} = nC_p\left(T_1-T_2\right) \;(VIII)$
+ </center>
+ <p class="TxtSolutions">
+ Substituting $(VII)$ and $(VIII)$ into $(III)$
+ </p>
+ <center>
+ $Q_c = nRT_1\ln{\frac{P_1}{P_2}}+\frac{i+2}{2}nR\left(T_1-T_2\right) \;(IX)$
+ </center>
+ <p class="TxtSolutions">
+ Substituting $(VI)$ and $(IX)$ into $(I)$ and developing algebraically
+ </p>
+ <center>
+ $\eta = \frac{2\left(T_2-T_1\right)\ln{\frac{P_2}{P_1}}}{(i+2)\left(T_2-T_1\right)+2T_2\ln{\frac{P_2}{P_1}}}$
+ </center>
+ <p class="TxtSolutions">
+ Since gas is monoatomic, $i = 3$, hence
+ </p>
+ <center>
+ $\boxed{\eta = \frac{2\left(T_2-T_1\right)\ln{\frac{P_2}{P_1}}}{5\left(T_2-T_1\right)+2T_2\ln{\frac{P_2}{P_1}}}}$
+ </center>
+ <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;">
+ BSc. Luis Daniel Fernández Quintana<br>
+ Physics Department (FCNE)<br>
+ Universidad de Oriente, Cuba<br>
+ </p>
+ </article>
+ </main>
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