Updated greek laters @ latex compiling

astrosander edited
revision #10171 parent #9179 GitHub dbb31ba ← older newer →
@@ -67,13 +67,13 @@
<h3>Solution</h3>
<p>
Let denote left volume as $V_1$ and right one as $V_2$. Applying state equation for each container
− $$P_1~V_1 = nRT \;(1)$$
− $$P_0~V_2 = nRT_0 \;(2)$$
+ $$P_1~V_1 = nRT \quad(1)$$
+ $$P_0~V_2 = nRT_0 \quad(2)$$
It is easy to see that
− $$V_1 = V_0+S\frac{l}{2}-Sx \;(3)$$
− $$V_2 = V_0+S\frac{l}{2}+Sx \;(4)$$
+ $$V_1 = V_0+S\frac{l}{2}-Sx \quad(3)$$
+ $$V_2 = V_0+S\frac{l}{2}+Sx \quad(4)$$
If droplet is in equilibrium that means pressure forces due to gasses are equal, so $P_1 = P_0$. Then, dividing $(1)$ and $(2)$
− $$\frac{V_1}{V_2}=\frac{T}{T_0} \;(5)$$
+ $$\frac{V_1}{V_2}=\frac{T}{T_0} \quad(5)$$
Substituting $(3)$ and $(4)$ into $(5)$ and separating $T$
$$\boxed{T = T_0~\frac{2V_0+S(l+2x)}{2V_0+S(l-2x)}}$$
This is mathematically valid for $0< x <\frac{V_0}{S}+\frac{l}{2}$ but according the physical situation only for $0< x < \frac{l}{2}$.
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