Capillary Rise in Closed Tube Solution

Solution: Capillary Rise in a Closed Tube

Problem Statement: A glass capillary of length $l = 1.01 \text{ m}$ is closed at the top and dipped in water. We need to find the height $h$ the water rises to, given that in an open tube it rises to $h_0 = 1.10 \text{ cm}$.

Given Data:

Length of tube, $l = 1.01 \text{ m}$
Capillary rise in open tube, $h_0 = 1.10 \text{ cm} = 0.011 \text{ m}$
Atmospheric Pressure, $P_0 = 1.01 \times 10^5 \text{ Pa}$
Density of water, $\rho = 1000 \text{ kg/m}^3$
Gravity, $g = 10 \text{ m/s}^2$

Step 1: Capillary Pressure Excess

In an open tube, the water rises to height $h_0$ due to surface tension. The pressure difference across the meniscus that supports this weight is:

$$ \Delta P_{surface\_tension} = \frac{2S}{r} = \rho g h_0 $$

This relationship $\frac{2S}{r} = \rho g h_0$ holds true regardless of the air pressure above the meniscus.

Step 2: Analysis of Trapped Air

Since the tube is closed at the top, as water rises to height $h$, the air inside is compressed. We assume the process is isothermal.

Initial State: Volume $V_1 \propto l$, Pressure $P_1 = P_0$
Final State: Volume $V_2 \propto (l – h)$, Pressure $P’$

Using Boyle’s Law ($P_1 V_1 = P_2 V_2$):

$$ P_0 \cdot l = P’ \cdot (l – h) \implies P’ = P_0 \frac{l}{l – h} $$

Step 3: Pressure Balance Equation

Consider the pressure just below the meniscus (point inside water). The pressure inside the water at the level of the reservoir surface is $P_0$. Going up by height $h$ to the meniscus, the pressure decreases by $\rho g h$.

$$ P_{water\_below\_meniscus} = P_0 – \rho g h $$

Across the curved meniscus, the pressure on the concave side (air) is higher than the convex side (water) by $\frac{2S}{r}$.

$$ P’ – P_{water\_below\_meniscus} = \frac{2S}{r} $$

Substituting the expressions we found:

$$ P’ – (P_0 – \rho g h) = \rho g h_0 $$ $$ P’ = P_0 + \rho g h_0 – \rho g h $$

Step 4: Solving for h

Equate the two expressions for $P’$:

$$ P_0 \frac{l}{l – h} = P_0 + \rho g (h_0 – h) $$

Rearranging to solve for $h$:

$$ \frac{P_0 l}{l – h} – P_0 = \rho g (h_0 – h) $$ $$ P_0 \left( \frac{l}{l – h} – 1 \right) = \rho g (h_0 – h) $$ $$ P_0 \left( \frac{h}{l – h} \right) = \rho g (h_0 – h) $$ $$ \frac{P_0}{\rho g} \left( \frac{h}{l – h} \right) = h_0 – h $$

Exact vs. Approximation:
Strictly speaking, this is a quadratic equation in $h$. However, because $l = 1.01 \text{ m}$ and $h$ is expected to be very small (in mm range), $l – h \approx l$.
Using this valid approximation ($l-h \approx l$) simplifies the algebra significantly without losing precision for the given significant figures.

Using the approximation $l – h \approx l$:

$$ \frac{P_0}{\rho g} \left( \frac{h}{l} \right) = h_0 – h $$ $$ h \left( \frac{P_0}{\rho g l} \right) + h = h_0 $$ $$ h \left( 1 + \frac{P_0}{\rho g l} \right) = h_0 $$

$$ h = \frac{h_0}{1 + \frac{P_0}{\rho g l}} $$

Step 5: Calculation

Substitute the values:

$$ h = \frac{0.011}{1 + \frac{1.01 \times 10^5}{1000 \times 10 \times 1.01}} $$

Simplify the denominator term:

$$ \frac{1.01 \times 10^5}{10^4 \times 1.01} = \frac{10^5}{10^4} = 10 $$

Final calculation:

$$ h = \frac{0.011}{1 + 10} = \frac{0.011}{11} = 0.001 \text{ m} $$ $$ h = 1.0 \text{ mm} $$

Answer: The water rises to a height of 1.0 mm.