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+ <header style="text-align:center;">
<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>
+ <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+
+ <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"
@@ -45,95 +57,38 @@
</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">
+
+ <h3>Solution</h3>
+ <p>
+ 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
+ $$\eta = 1-\frac{|Q_c|}{Q_h}=1-\frac{|Q_{34}|}{Q_{12}} \;(1)$$
+ 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,
+ $$Q_{34}=\Delta U_{34} = nC_v\left(T_4-T_3\right) \;(2)$$
From Poisson Law
− </p>
− <center>
− $P_1V_1^\gamma = P_4V_2^\gamma$
− </center>
− <p class="TxtSolutions">
+ $$P_1V_1^\gamma = P_4V_2^\gamma$$
and the state 4 is characterized by
− </p>
− <center>
− $P_4V_2=nRT_4$
− </center>
− <p class="TxtSolutions">
+ $$P_4V_2=nRT_4$$
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">
+ $$T_4 = \frac{P_1V_1^\gamma}{nRV_2^{\gamma-1}} \;(3)$$
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">
+ $$P_2V_1^\gamma=P_3V_2^\gamma$$
State equation for state 3 is
− </p>
− <center>
− $P_3V_2=nRT_3$
− </center>
− <p class="TxtSolutions">
+ $$P_3V_2=nRT_3$$
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">
+ $$T_3 = \frac{P_2V_1^\gamma}{nRV_2^{\gamma-1}} \;(4)$$
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">
+ $$Q_{34} = \frac{i}{2}\frac{V_1^\gamma}{V_2^{\gamma-1}}\left(P_1-P_2\right) \;(5)$$
For the process $1-2$
− </p>
− <center>
− $Q_{12}=\Delta U_{12} = nC_v\left(T_2-T_1\right)$
− </center>
− <p class="TxtSolutions">
+ $$Q_{12}=\Delta U_{12} = nC_v\left(T_2-T_1\right)$$
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">
+ $$Q_{12}=nC_v\left(\frac{P_2V_1}{nR}-\frac{P_1V_1}{nR}\right)$$
+ $$Q_{12}=\frac{i}{2}V_1\left(P_2-P_1\right) \;(6)$$
Putting $(5)$ and $(6)$ into $(1)$
− </p>
− <center>
− $\eta = 1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}$
− </center>
− <p class="TxtSolutions">
+ $$\eta = 1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$
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">
+ $$\boxed{\eta = 1-\left(\frac{V_1}{V_2}\right)^{\frac{2}{3}}}$$
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">
+ $$\eta = 1-\frac{|Q_c|}{Q_h} \;(I)$$
For identifying correctly the reservoirs, let's see the following graph corresponding to the cycle in a $P$ vs $V$ plot.
</p>
<center>
@@ -145,113 +100,50 @@
</figcaption>
</figure>
</center>
− <p class="TxtSolutions">
+ <p>
The hot reservoir is defined as
− </p>
− <center>
− $Q_h = Q_{23} + Q_{34} \;(II)$
− </center>
− <p class="TxtSolutions">
+ $$Q_h = Q_{23} + Q_{34} \;(II)$$
and cold one
− </p>
− <center>
− $Q_c = Q_{12} + Q_{41} \;(III)$
− </center>
− <p class="TxtSolutions">
+ $$Q_c = Q_{12} + Q_{41} \;(III)$$
Process $2-3$ is isobaric, then
− </p>
− <center>
− $Q_{23}=nC_p\left(T_2-T_1\right) \;(IV)$
− </center>
− <p class="TxtSolutions">
+ $$Q_{23}=nC_p\left(T_2-T_1\right) \;(IV)$$
Process $3-4$ is isothermal, so
− </p>
− <center>
− $Q_{34} = A_{34} = nRT_2\ln{\frac{V_4}{V_3}}$
− </center>
− <p class="TxtSolutions">
+ $$Q_{34} = A_{34} = nRT_2\ln{\frac{V_4}{V_3}}$$
Since $V=nRT/P$ according state equation
− </p>
− <center>
− $Q_{34} = nRT_2\ln{\frac{P_2}{P_1}} \;(V)$
− </center>
− <p class="TxtSolutions">
+ $$Q_{34} = nRT_2\ln{\frac{P_2}{P_1}} \;(V)$$
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">
+ $$Q_h = \frac{i+2}{2}nR\left(T_2-T_1\right)+nRT_2\ln{\frac{P_2}{P_1}} \;(VI)$$
Process $1-2$ is isothermal, so
− </p>
− <center>
− $Q_{12}=A_{12}=nRT_1\ln{\frac{V_2}{V_1}}$
− </center>
− <p class="TxtSolutions">
+ $$Q_{12}=A_{12}=nRT_1\ln{\frac{V_2}{V_1}}$$
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">
+ $$Q_{12}=A_{12}=nRT_1\ln{\frac{P_1}{P_2} \;(VII)}$$
Finally, process $4-1$ is isobaric
− </p>
− <center>
− $Q_{14} = nC_p\left(T_1-T_2\right) \;(VIII)$
− </center>
− <p class="TxtSolutions">
+ $$Q_{14} = nC_p\left(T_1-T_2\right) \;(VIII)$$
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">
+ $$Q_c = nRT_1\ln{\frac{P_1}{P_2}}+\frac{i+2}{2}nR\left(T_1-T_2\right) \;(IX)$$
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">
+ $$\eta = \frac{2\left(T_2-T_1\right)\ln{\frac{P_2}{P_1}}}{(i+2)\left(T_2-T_1\right)+2T_2\ln{\frac{P_2}{P_1}}}$$
Since gas is monoatomic, $i = 3$, hence
+ $$\boxed{\eta = \frac{2\left(T_2-T_1\right)\ln{\frac{P_2}{P_1}}}{5\left(T_2-T_1\right)+2T_2\ln{\frac{P_2}{P_1}}}}$$
</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>
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+ BSc. Luis Daniel Fernández Quintana<br>
+ Physics Department (FCNE)<br>
+ Universidad de Oriente, Cuba<br>
+ </p>
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