Problem 24 - A Modern Course in Statistical Physics

Question

Compute the equilibrium vapor pressure of a monomolecular gas in equilibrium with, a spherical droplet of liquid of the same substance, as a function of the radius \(R\) of the droplet and for fixed temperature. Assume the gas phase is well described by the ideal gas equation of state and the liquid can be assumed to be incompressible. Use the fact that for mechanical equilibrium \(P_{1}-P_{\mathrm{g}}=\) \(2 \sigma / R\), where \(P_{1}\left(P_{g}\right)\) is the pressure of the liquid (gas) and \(\sigma\) is the surface tension.

Step-by-Step Solution

Answer

The equilibrium vapor pressure is given by \[ P_{g} = P_{1} - \frac{2 \sigma}{R} \].

Step 1 - Analyzing the Given Relationship

The given relationship for mechanical equilibrium is \[ P_{1} - P_{\text{g}} = \frac{2 \sigma}{R} \] where \(P_{1}\) is the pressure of the liquid, \(P_{g}\) is the pressure of the gas, \(\sigma\) is the surface tension, and \(R\) is the radius of the droplet.

Step 2 - Express Gas Pressure with Ideal Gas Law

Since the gas phase is well described by the ideal gas equation of state, we can use \[ P_{\text{g}} V = nRT \] where \(P_{\text{g}}\) is the gas pressure, \(V\) is the volume, \(n\) is the number of moles of gas, \(R\) is the gas constant, and \(T\) is the temperature. Rearranging this, we get \[ P_{\text{g}} = \frac{nRT}{V} \].

Step 3 - Volume of the Droplet

The volume \(V\) of a spherical droplet is given by \[ V = \frac{4}{3} \left ( \pi R^3 \right )\].

Step 4 - Equilibrium Vapor Pressure Equation

Given the mechanical equilibrium relationship \( P_{1} - P_{\text{g}} = \frac{2 \sigma}{R} \), substitute \(P_{\text{g}}\) with \[ P_{\text{g}} = \frac{nRT}{\frac{4}{3} \pi R^3} \] to get: \[ P_{1} - \frac{nRT}{\frac{4}{3} \pi R^3} = \frac{2 \sigma}{R} \].

Step 5 - Isolate Equilibrium Vapor Pressure

To isolate the equilibrium vapor pressure \(P_{g}\), rearrange the equation: \[ P_{g} = P_{1} - \frac{2 \sigma}{R} \].

Step 6 - Final Equilibrium Vapor Pressure

Thus, the equilibrium vapor pressure of the monomolecular gas in equilibrium with the spherical droplet of liquid of the same substance is \[ P_{g} = P_{1} - \frac{2 \sigma}{R} \]. This shows that the equilibrium vapor pressure decreases as the radius of the droplet decreases.

Key Concepts

mechanical equilibrium

In physics, mechanical equilibrium refers to a state where there are no unbalanced forces acting on a system, meaning the system is either at rest or moving with constant velocity. For a spherical droplet in equilibrium, this means the pressure inside the droplet must balance with the external pressure plus the effect of surface tension.
The given relationship for mechanical equilibrium is described by:
\[ P_{1} - P_{g} = \frac{2 \sigma}{R} \]
Here,

• \(P_{1}\) is the liquid pressure inside the droplet,
  
• \(P_{g}\) is the gas pressure outside the droplet,
  
• \(\sigma\) is the surface tension of the liquid,
  
• and \(R\) is the radius of the droplet.

This equation shows that the pressure difference across the liquid-gas interface is proportional to the surface tension and inversely proportional to the radius of the droplet.

ideal gas law

The ideal gas law is a fundamental equation that describes the behavior of an ideal gas. It is given by:
\[ P_{g} V = nRT \]
where,

• \(P_{g}\) is the gas pressure,
  
• \(V\) is the volume,
  
• \(n\) is the number of moles,
  
• \(R\) is the gas constant, and
• \(T\) is the temperature.

This equation helps us express the gas pressure in terms of the number of moles, the gas constant, the temperature, and the volume. By rearranging this formula, we get:
\[ P_{g} = \frac{nRT}{V} \]. This expression allows us to relate the gas pressure to its volume and other parameters. In the context of our problem, it helps to substitute \(P_{g}\) in the equilibrium equation to solve for the equilibrium vapor pressure.

surface tension

Surface tension is a physical property of liquids that causes them to behave as if their surface were covered with a stretched elastic membrane. It arises due to the attraction forces between the molecules at the surface of the liquid. Surface tension is crucial in determining the shape of liquid droplets and bubbles.
In our equation, surface tension plays a vital role in the pressure difference across the liquid-gas interface of the droplet, represented by \(2 \sigma / R\). This term shows that the surface tension's effect is more pronounced for smaller droplets (smaller \(R\)). Hence, as the radius \(R\) of the droplet decreases, the contribution of surface tension to the pressure difference becomes more significant.

spherical droplet

A spherical droplet is a perfectly round liquid droplet with uniform curvature. The volume of a spherical object is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
where \(R\) is the radius of the sphere.
In the context of our problem, this formula is essential for determining the volume of the droplet. When substituted into the ideal gas law, it links the gas pressure with the droplet's radius. This allows us to understand how the radius impacts the gas pressure and, subsequently, the equilibrium vapor pressure. For instance, if the volume \(V\) was incorporated into the ideal gas law, substituting \(V\) as \( \frac{4}{3} \pi R^3 \) aids in solving for \( P_{g} \).