Updated greek laters @ latex compiling

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@@ -67,34 +67,34 @@
<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
− $$\eta = 1-\frac{|Q_c|}{Q_h}=1-\frac{|Q_{34}|}{Q_{12}} \;(1)$$
+ $$\eta = 1-\frac{|Q_c|}{Q_h}=1-\frac{|Q_{34}|}{Q_{12}} \quad(1)$$
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,
− $$Q_{34}=\Delta U_{34} = nC_v\left(T_4-T_3\right) \;(2)$$
+ $$Q_{34}=\Delta U_{34} = nC_v\left(T_4-T_3\right) \quad(2)$$
From Poisson Law
$$P_1V_1^\gamma = P_4V_2^\gamma$$
and the state 4 is characterized by
$$P_4V_2=nRT_4$$
So, separating $P_4$ and putting it into above Poisson law
− $$T_4 = \frac{P_1V_1^\gamma}{nRV_2^{\gamma-1}} \;(3)$$
+ $$T_4 = \frac{P_1V_1^\gamma}{nRV_2^{\gamma-1}} \quad(3)$$
Proceeding similarly, $T_3$ can be found. Applying Poisson Law for process $2-3$,
$$P_2V_1^\gamma=P_3V_2^\gamma$$
State equation for state 3 is
$$P_3V_2=nRT_3$$
Separating $P_3$ and substituting into respective Poisson Law
− $$T_3 = \frac{P_2V_1^\gamma}{nRV_2^{\gamma-1}} \;(4)$$
+ $$T_3 = \frac{P_2V_1^\gamma}{nRV_2^{\gamma-1}} \quad(4)$$
Substituting $(3)$ and $(4)$ into $(2)$, and as $C_v=iR/2$
− $$Q_{34} = \frac{i}{2}\frac{V_1^\gamma}{V_2^{\gamma-1}}\left(P_1-P_2\right) \;(5)$$
+ $$Q_{34} = \frac{i}{2}\frac{V_1^\gamma}{V_2^{\gamma-1}}\left(P_1-P_2\right) \quad(5)$$
For the process $1-2$
$$Q_{12}=\Delta U_{12} = nC_v\left(T_2-T_1\right)$$
Taking account the state equations for states 1 and 2, temperatures can be found,
$$Q_{12}=nC_v\left(\frac{P_2V_1}{nR}-\frac{P_1V_1}{nR}\right)$$
− $$Q_{12}=\frac{i}{2}V_1\left(P_2-P_1\right) \;(6)$$
+ $$Q_{12}=\frac{i}{2}V_1\left(P_2-P_1\right) \quad(6)$$
Putting $(5)$ and $(6)$ into $(1)$
$$\eta = 1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$
Since $\gamma=C_p/C_v=1+2/i$ and $i=3$ (monoatomic gas),
$$\boxed{\eta = 1-\left(\frac{V_1}{V_2}\right)^{\frac{2}{3}}}$$
b) Let enumerate again the states following a clockwise way, this time starting in coordinate $(P_1,T_1)$. The efficience is
− $$\eta = 1-\frac{|Q_c|}{Q_h} \;(I)$$
+ $$\eta = 1-\frac{|Q_c|}{Q_h} \quad(I)$$
For identifying correctly the reservoirs, let's see the following graph corresponding to the cycle in a $P$ vs $V$ plot.
</p>
<center>
@@ -108,25 +108,25 @@
</center>
<p>
The hot reservoir is defined as
− $$Q_h = Q_{23} + Q_{34} \;(II)$$
+ $$Q_h = Q_{23} + Q_{34} \quad(II)$$
and cold one
− $$Q_c = Q_{12} + Q_{41} \;(III)$$
+ $$Q_c = Q_{12} + Q_{41} \quad(III)$$
Process $2-3$ is isobaric, then
− $$Q_{23}=nC_p\left(T_2-T_1\right) \;(IV)$$
+ $$Q_{23}=nC_p\left(T_2-T_1\right) \quad(IV)$$
Process $3-4$ is isothermal, so
$$Q_{34} = A_{34} = nRT_2\ln{\frac{V_4}{V_3}}$$
Since $V=nRT/P$ according state equation
− $$Q_{34} = nRT_2\ln{\frac{P_2}{P_1}} \;(V)$$
+ $$Q_{34} = nRT_2\ln{\frac{P_2}{P_1}} \quad(V)$$
Substituting $(IV)$ and $(V)$ into $(II)$and taking account that $C_p=(i+2)R/2$
− $$Q_h = \frac{i+2}{2}nR\left(T_2-T_1\right)+nRT_2\ln{\frac{P_2}{P_1}} \;(VI)$$
+ $$Q_h = \frac{i+2}{2}nR\left(T_2-T_1\right)+nRT_2\ln{\frac{P_2}{P_1}} \quad(VI)$$
Process $1-2$ is isothermal, so
$$Q_{12}=A_{12}=nRT_1\ln{\frac{V_2}{V_1}}$$
and newly, taking account state equation
− $$Q_{12}=A_{12}=nRT_1\ln{\frac{P_1}{P_2} \;(VII)}$$
+ $$Q_{12}=A_{12}=nRT_1\ln{\frac{P_1}{P_2} \quad(VII)}$$
Finally, process $4-1$ is isobaric
− $$Q_{14} = nC_p\left(T_1-T_2\right) \;(VIII)$$
+ $$Q_{14} = nC_p\left(T_1-T_2\right) \quad(VIII)$$
Substituting $(VII)$ and $(VIII)$ into $(III)$
− $$Q_c = nRT_1\ln{\frac{P_1}{P_2}}+\frac{i+2}{2}nR\left(T_1-T_2\right) \;(IX)$$
+ $$Q_c = nRT_1\ln{\frac{P_1}{P_2}}+\frac{i+2}{2}nR\left(T_1-T_2\right) \quad(IX)$$
Substituting $(VI)$ and $(IX)$ into $(I)$ and developing algebraically
$$\eta = \frac{2\left(T_2-T_1\right)\ln{\frac{P_2}{P_1}}}{(i+2)\left(T_2-T_1\right)+2T_2\ln{\frac{P_2}{P_1}}}$$
Since gas is monoatomic, $i = 3$, hence
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