Calculate Capillary Rise in Isothermally Compressed Container

To find the height $h$ of the liquid column in the capillary tube above the liquid level, we analyze the pressure equilibrium at the liquid surface inside the container.

Step 1: Compression of Air in the Container

Initially, the air inside the container is at atmospheric pressure $P_0 = 10^5 \text{ N m}^{-2}$ and occupies a volume $V_0$.

The air is isothermally compressed to a final volume $V' = \frac{100}{101}V_0$. Using Boyle's Law ($P_1 V_1 = P_2 V_2$) for an isothermal process: $P_0 V_0 = P_{\text{air}} \left(\frac{100}{101}V_0\right)$

Solving for the new air pressure $P_{\text{air}}$ inside the container: $P_{\text{air}} = \frac{101}{100} P_0 = 1.01 P_0$

Thus, the excess air pressure inside the container above atmospheric pressure is: $P_{\text{air}} - P_0 = 0.01 P_0 = 0.01 \times 10^5 \text{ N m}^{-2} = 1000 \text{ N m}^{-2}$

---

Step 2: Pressure Balance in the Capillary Tube

The top end of the capillary tube is open to the atmosphere, so the pressure at the surface of the meniscus inside the capillary tube is atmospheric pressure $P_0$.

Due to surface tension and a contact angle of $\theta = 0^\circ$, the meniscus is concave upwards. The pressure drop across the spherical liquid meniscus is given by: $\Delta P_{\text{surface tension}} = \frac{2T}{r}$

Given:

• Surface tension, $T = 0.075 \text{ N m}^{-1}$
• Capillary inner radius, $r = 0.1 \text{ mm} = 10^{-4} \text{ m}$

Calculating the pressure drop across the meniscus: $\Delta P_{\text{surface tension}} = \frac{2 \times 0.075}{10^{-4}} = 1500 \text{ N m}^{-2}$

Therefore, the pressure just below the liquid meniscus inside the capillary is: $P_{\text{below meniscus}} = P_0 - \frac{2T}{r}$

The pressure at the level of the liquid surface inside the container, but inside the capillary tube, is given by adding the hydrostatic pressure of the liquid column of height $h$: $P_{\text{capillary, level}} = P_0 - \frac{2T}{r} + \rho g h$

---

Step 3: Solving for the Column Height $h$

By hydrostatic equilibrium, the pressure inside the capillary tube at the liquid level must equal the air pressure pressing on the liquid surface outside the capillary: $P_{\text{air}} = P_0 - \frac{2T}{r} + \rho g h$

Rearranging the terms: $P_{\text{air}} - P_0 + \frac{2T}{r} = \rho g h$

Substitute the given and calculated values:

• $P_{\text{air}} - P_0 = 1000 \text{ N m}^{-2}$
• $\frac{2T}{r} = 1500 \text{ N m}^{-2}$
• $\rho = 10^3 \text{ kg m}^{-3}$
• $g = 10 \text{ m s}^{-2}$

$1000 + 1500 = (10^3 \times 10) h$ $2500 = 10^4 h$ $h = \frac{2500}{10^4} = 0.25 \text{ m}$

Converting $h$ into centimeters: $h = 0.25 \times 100 \text{ cm} = 25 \text{ cm}$

Final Answer: The height $h$ of the liquid column in the capillary above the liquid level is 25.