We have two isochoric processes, 1-2 and 3-4 ($V_1=V_2$ and $V_3=V_4$). And two isobaric processes, 2-3 and 4-1 ($P_2=P_3$ and $P_1=P_4), and we know that $T_2=T_4$\
+
We have two isochoric processes, 1-2 and 3-4 ($V_1=V_2$ and $V_3=V_4$). And two isobaric processes, 2-3 and 4-1 ($P_2=P_3$ and $P_1=P_4$), and we know that $T_2=T_4$\
$W_{12}=0$, $W_{34}=0$ (isochoric processes)\
so, the total work will be\
$W_T=W_{23} +W_{41}$\
$W_T=P_2(V_3 -V_2)+P_1(V_1-V_4)$
$W_T=P_2V_3 -P_2V_2 +P_1V_1 -P_1V_4$\
−
but $P_2V_3=RT_3$, $P_1V_1=RT_1$, and $P_2V_2=P_4V_4=RT_2=RT_4 (because $T_2=T_4$)\
+
but $P_2V_3=RT_3$, $P_1V_1=RT_1$, and $P_2V_2=P_4V_4=RT_2=RT_4$ (because $T_2=T_4$)\
and from this relations, we get in the work\
$W_T=RT_3 +RT_1 -2RT_2$\
From Gay-Lussac's law\
−
$\frac{P_1}{T_1}=\frac{P_2}{T_2}$, and $\frac{P_3}{T_3}=\frac{P_4}{T_4$}$\
+
$\frac{P_1}{T_1}=\frac{P_2}{T_2}$, and $\frac{P_3}{T_3}=\frac{P_4}{T_4}$\
Multiplying term by term\
$\frac{P_1P_3}{T_1T_3}=\frac{P_2P_4}{T_2T_4}$\
we get\
$\frac{1}{T_1T_3}=\frac{1}{(T_2)^2}$\
so $T_2=(T_1T_3)^\frac{1}{2}$\
and susbtituting this into the total work\
@@ -26,8 +30,8 @@Solution
$W_T=RT_3 +RT_1 -2R(T_1T_3)^\frac{1}{2}$\
$W_T=R[T_3 + T_1 -2(T_1T_3)^\frac{1}{2}]$\
so
−
$W_T=R[(T_3)^\frac{1}{2} -(T_1)^\frac{1}{2}]$
+
$W_T=R[(T_3)^\frac{1}{2} -(T_1)^\frac{1}{2}]^2$
#### Answer
−
[Insert a concise answer or boxed result]
+
$W_T=R[(T_3)^\frac{1}{2} -(T_1)^\frac{1}{2}]^2$
@@ -1,22 +1,26 @@
### Statement
### Statement
$5.6.18.$[Insert the problem statement]
$5.6.18.$ One mole of gas participates in a cyclic process, the graph of which, consisting
of two isochores and two isobars, is shown in the figure. The temperature at
points 1 and 3 is T1 and T3. Determine the work done by the gas per cycle if it
is known that points 2 and 4 lie on the same isotherm.
We have two isochoric processes, 1-2 and 3-4 ($V_1=V_2$ and $V_3=V_4$). And two isobaric processes, 2-3 and 4-1 ($P_2=P_3$ and $P_1=P_4), and we know that $T_2=T_4$\
We have two isochoric processes, 1-2 and 3-4 ($V_1=V_2$ and $V_3=V_4$). And two isobaric processes, 2-3 and 4-1 ($P_2=P_3$ and $P_1=P_4$), and we know that $T_2=T_4$\
$W_{12}=0$, $W_{34}=0$ (isochoric processes)\
$W_{12}=0$, $W_{34}=0$ (isochoric processes)\
so, the total work will be\
so, the total work will be\
$W_T=W_{23} +W_{41}$\
$W_T=W_{23} +W_{41}$\
$W_T=P_2(V_3 -V_2)+P_1(V_1-V_4)$
$W_T=P_2(V_3 -V_2)+P_1(V_1-V_4)$
$W_T=P_2V_3 -P_2V_2 +P_1V_1 -P_1V_4$\
$W_T=P_2V_3 -P_2V_2 +P_1V_1 -P_1V_4$\
but $P_2V_3=RT_3$, $P_1V_1=RT_1$, and $P_2V_2=P_4V_4=RT_2=RT_4 (because $T_2=T_4$)\
but $P_2V_3=RT_3$, $P_1V_1=RT_1$, and $P_2V_2=P_4V_4=RT_2=RT_4$ (because $T_2=T_4$)\
and from this relations, we get in the work\
and from this relations, we get in the work\
$W_T=RT_3 +RT_1 -2RT_2$\
$W_T=RT_3 +RT_1 -2RT_2$\
From Gay-Lussac's law\
From Gay-Lussac's law\
$\frac{P_1}{T_1}=\frac{P_2}{T_2}$, and $\frac{P_3}{T_3}=\frac{P_4}{T_4$}$\
$\frac{P_1}{T_1}=\frac{P_2}{T_2}$, and $\frac{P_3}{T_3}=\frac{P_4}{T_4}$\
