Problem 28 - A Modern Course in Statistical Physics

Question

A tiny sack made of membrane permeable to water but not \(\mathrm{NaCl}\) (sodium chloride) is filled with a \(1 \%\) solution (by weight) of \(\mathrm{NaCl}\) and water and is immersed in an open beaker of pure water at \(38{ }^{\circ} \mathrm{C}\) at a depth of \(1 \mathrm{ft}\). (a) What osmotic pressure is experienced by the sack? (b) What is the total pressure of the solution in the sack (neglect surface tension)? Assume that the sack is small enough that the pressure of the surrounding water can be assumed constant. (An example of such a sack is a human blood cell.)

Step-by-Step Solution

Answer

(a) The osmotic pressure is 8.73 atm. (b) The total pressure is 8.76 atm.

Step 1 - Calculate Solute Concentration

The solution within the sack is given as a 1% NaCl solution by weight. That means 1 gram of NaCl is dissolved in 100 grams of water. The molar mass of NaCl is approximately 58.44 g/mol. Therefore, the molar concentration (C) can be calculated by dividing the mass of the solute by its molar mass and the volume of the solution.

Step 2 - Find Osmotic Pressure

Using the formula for osmotic pressure \[ \text{Osmotic Pressure} (\text{π}) = iCRT \] where \(i\) is the van 't Hoff factor for NaCl (which is 2, since NaCl dissociates into 2 ions, Na⁺ and Cl⁻), \(C\) is the molar concentration calculated in Step 1, \(R\) is the gas constant (0.0821 L atm / K mol), and \(T\) is the temperature in Kelvins. Convert the temperature from Celsius to Kelvin using the formula \(T(K) = T(°C) + 273.15\).

Step 3 - Calculate Molar Concentration

The molar concentration of NaCl: \[ C = \frac{1 \text{ g}}{58.44 \text{ g/mol} \times 0.1 \text{ L}} = 0.171 \text{ mol/L} \] Now use this concentration to find the osmotic pressure.

Step 4 - Calculate Temperature in Kelvins

Convert the temperature to Kelvins: \[ 38°C + 273.15 = 311.15 \text{ K} \]

Step 5 - Apply Osmotic Pressure Formula

Substitute the values into the osmotic pressure formula: \[ \text{π} = 2 \times 0.171 \times 0.0821 \times 311.15 \] Calculate the result to get the osmotic pressure.

Step 6 - Calculate Osmotic Pressure

Perform the multiplication: \[ \text{π} ≈ 8.73 \text{ atm} \]

Step 7 - Find Hydrostatic Pressure

The hydrostatic pressure due to the depth of the water is given by \[ P = \rho gh \] where \(\rho\) is the density of water, \(g\) is the acceleration due to gravity, and \(h\) is the depth. For water, \(\rho ≈ 1000 \text{ kg/m}^3\), \(g ≈ 9.81 \text{ m/s}^2\), and \(h = 1 \text{ ft} = 0.3048 \text{ m}\).

Step 8 - Calculate Hydrostatic Pressure

Substitute values into the hydrostatic pressure formula: \[ P = 1000 \times 9.81 \times 0.3048 \] Calculate the result to find the pressure.

Step 9 - Calculate Pressure Due to Depth

\[ P ≈ 2.98 \text{ kPa} \text{ or } 0.0293 \text{ atm} \]

Step 10 - Total Pressure Calculation

The total pressure in the solution is the sum of the osmotic pressure and the hydrostatic pressure: \[ \text{Total Pressure} = \text{Osmotic Pressure} + \text{Hydrostatic Pressure} \]

Step 11 - Add Pressures

\[ \text{Total Pressure} = 8.73 \text{ atm} + 0.0293 \text{ atm} = 8.76 \text{ atm} \]

Key Concepts

Osmotic Pressure

Osmotic pressure is the pressure required to stop the flow of water through a semi-permeable membrane that separates two solutions with different concentrations. It occurs because water tends to move from an area of low solute concentration to an area of high solute concentration, trying to equalize concentrations on both sides of the membrane. This pressure is vital in many biological processes, such as the maintenance of proper cell turgor in plants and the function of kidneys in animals.
In the given exercise, the osmotic pressure experienced by the sack filled with 1% NaCl solution in pure water can be calculated using the formula:
\[ \text{Osmotic Pressure} (\pi) = iCRT \]
Here, \(i\) is the van 't Hoff factor (2 for NaCl as it dissociates into Na⁺ and Cl⁻ ions), \(C\) is the molar concentration of the solution, \(R\) is the gas constant (0.0821 L atm / K mol), and \(T\) is the temperature in Kelvins. This relationship helps us understand the pressure developed due to the osmosis process.

Molar Concentration

Molar concentration (C), also known as molarity, is the number of moles of a solute dissolved in one liter of solution. It's a standard way to express concentrations in chemistry because it directly relates the amount of solute to the volume of a solution, making it easier to calculate and understand reactions and properties.
To find the molar concentration of NaCl in the solution, follow these steps:

• Find the mass of NaCl: 1 gram
• Convert mass to moles using the molar mass of NaCl (58.44 g/mol):

\[ \text{Moles of NaCl} = \frac{1 \text{ g}}{58.44 \text{ g/mol}} = 0.0171 \text{ mol} \]

• Assume the volume of the solution is approximately 0.1 liters (for convenience, 100 grams of water is roughly 100 mL):

\[ \text{Molar Concentration} (C) = \frac{0.0171 \text{ mol}}{0.1 \text{ L}} = 0.171 \text{ mol/L} \]
This molar concentration value is used in the osmotic pressure calculation.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It's a crucial concept when dealing with fluids at rest, such as water in a container or blood in vessels.
Hydrostatic pressure can be calculated using the formula:
\[ P = \rho gh \]
where \(\rho\) is the fluid density, \(g\) is the acceleration due to gravity, and \(h\) is the depth.
For the given problem, with water at 1-foot depth:

• Density of water: 1000 kg/m³
• Acceleration due to gravity: 9.81 m/s²
• Depth: 0.3048 meters (since 1 ft = 0.3048 m)

Substitute these values into the formula:
\[ P = 1000 \times 9.81 \times 0.3048 = 2.98 \text{ kPa} \]
which is approximately 0.0293 atm. This pressure is due to the depth of the fluid and is added to the osmotic pressure to get the total pressure.

van 't Hoff Factor

The van 't Hoff factor (represented as \(i\) in the osmotic pressure formula) is a measure of the effect of a solute on the colligative properties of a solution. It reflects the number of particles into which a solute dissociates in solution.
For instance, NaCl disassociates into two ions (Na⁺ and Cl⁻) in water, so its van 't Hoff factor \(i\) is 2. This factor is crucial because it directly impacts the magnitude of colligative properties like osmotic pressure, boiling point elevation, and freezing point depression.
In the osmotic pressure equation \[ \pi = iCRT, \]
the van 't Hoff factor \(i\) multiplies the concentration of the solute, enhancing the effect since more particles contribute to the property. Accurate determination of \(i\) is essential for precise calculations and understanding the behavior of solutions, particularly electrolytes.