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GR9277 #46
Problem
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\prob{46}

Isotherms and coexistence curves are shown in the pV diagram above for a liquid-gas system. The dashed lines are the boundaries of the labeled regions.

Which numbered curve is the critical isotherm?

1. 1
2. 2
3. 3
4. 4
5. 5

Thermodynamics$\Rightarrow$}Critical Isotherm

The critical isotherm is the (constant temperature) line that just touches the critical liquid-vapor region, explained in the next question. The condition for the critical isotherm is $\left(\frac{dP}{dV}\right)_c=0$ and $\left(\frac{d^2P}{dV^2}\right)_c=0$, where c denotes the critical point.

Alternate Solutions
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neon37
2010-11-11 23:25:48
Easy way even without knowing anything in physics at all. There is nothing "critical" about 1, 3 -5. Only 2 looks "critical".
 droosenoose2019-10-18 19:49:04 that what I did lol
sirenayka
2008-10-14 21:49:48
Ning Bao
2008-01-29 11:48:28
Immediately we know that there is ino difference between 3, 4, and 5: their properties are the same. There is also nothing differentiating curve 1 from a curve just above or just below it. Only curve 2 is unique enough to be the anwer.
bucky0
2007-11-01 14:48:41
LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$