# Fluids and thermodynamics

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1. May 11, 2015

1. The problem statement, all variables and given/known data

2. Relevant equations

height of liquid above water level $h=\frac{2T}{R\rho g}$
for isothermal process :$PV$=constant
And if $P_0$ is atm. pressure, and P is pressure just below the water level in capillary tube, then $$P=P_0-\frac{2T}{R}$$
3. The attempt at a solution

Let y be the height of tube above water level. (beaker is open).
$P_1=P_0$
$V_1=\pi r^2(l-h)$
$V_2=\pi r^2y$
so from equation for isothermal process, final pressure of gas is
$P_2=\frac{P_0(l-h)}{y}$
the gas now exerts $P_2$ pressure on liquid inside tube.
So $P=\frac{P_0(l-h)}{y}-\frac{2T}{R_1}$ since the radius of curvature of liquid surface can change.

2. May 12, 2015

### Vibhor

I think you have approached the problem correctly . But there are too many P's floating around in your work :)

Isn't the answer given as $y = \frac{P_0(l-h)}{P_0+hρg}$ ?

Could you confirm the answer .

Last edited: May 12, 2015
3. May 13, 2015

4. May 13, 2015

$P_1=P_0$
$V_1=\pi r^2(l-h)$
$P_2=?$
$V_2=\pi r^2(y-h')$