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.)

Short Answer

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

Step by step solution

01

- 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.
02

- 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 \(T(K) = T(°C) + 273.15\).
03

- 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.
04

- Calculate Temperature in Kelvins

Convert the temperature to Kelvins: \[ 38°C + 273.15 = 311.15 \text{ K} \]
05

- 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.
06

- Calculate Osmotic Pressure

Perform the multiplication: \[ \text{π} ≈ 8.73 \text{ atm} \]
07

- 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 \(h = 1 \text{ ft} = 0.3048 \text{ m}\).
08

- 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.
09

- Calculate Pressure Due to Depth

\[ P ≈ 2.98 \text{ kPa} \text{ or } 0.0293 \text{ atm} \]
10

- Total Pressure Calculation

The total pressure in the solution is the sum of the osmotic pressure and the hydrostatic pressure: \[ \text{Total Pressure} = \text{Osmotic Pressure} + \text{Hydrostatic Pressure} \]
11

- Add Pressures

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two containers, each of volume \(V\), contain ideal gas held at temperature \(T\) and pressure \(P\). The gas in chamber 1 consists of \(N_{1, a}\) molecules of type \(a\) and \(N_{1, b}\) molecules of type \(b\). The gas in chamber 2 consists of \(N_{2, a}\) molecules of type \(a\) and \(N_{2, b}\) molecules of type \(b\). Assume that \(N_{1, a}+N_{1, b}=\) \(N_{2, a}+N_{2, b}\). The gases are allowed to mix so the final temperature is \(T\) and the final pressure is \(P\). (a) Compute the entropy of mixing. (b) What is the entropy of mixing if \(N_{1, \mathrm{a}}=N_{2, \mathrm{a}}\) and \(N_{1, \mathrm{~b}}=N_{2, \mathrm{~b}}\). (c) What is the entropy of mixing if \(N_{1, \mathrm{a}}=N_{2, \mathrm{~b}}\) and \(N_{1, \mathrm{~b}}=N_{2, \mathrm{a}}=0\). Discuss your results for (b) and (c).

Consider the reaction $$ 2 \mathrm{NH}_{3}=\mathrm{N}_{2}+3 \mathrm{H}_{2} $$ which occurs in the gas phase. Start initially with \(2 \mathrm{~mol}\) of \(\mathrm{NH}_{3}\) and 0 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{N}_{2}\). Assume that the reaction occurs at temperature \(T\) and pressure \(P\). Use ideal gas equations for the chemical potential. (a) Compute and plot the Gibbs free energy, \(G(T, P\), (5), as a function of the degree of reaction, \(\xi\), for (i) \(P=1\) atm and \(T=298 \mathrm{~K}\) and (ii) \(P=1 \mathrm{~atm}\) and \(T\). \(=894 \mathrm{~K}\). (b) Compute and plot the affinity, \(A(T, P, \xi)\), as a function of the degree of reaction, \(\xi\), for (i) \(P=1 \mathrm{~atm}\) and \(T=298 \mathrm{~K}\) and (ii) \(P=1\) atm and \(T=894 \mathrm{~K}\). (c) What is the degree of reaction, \(\xi\), at chemical equilibrium for \(P=1 a t m\) and temperature \(T=894\) K? How many moles of \(\mathrm{NH}_{3}, \mathrm{H}_{2}\), and \(\mathrm{N}_{2}\) are present at equilibrium? (d) If initially the volume is \(V_{0}\), what is the volume at equilibrium for \(P=\) \(1 \mathrm{~atm}\) and \(T=894 \mathrm{~K}\) ? (e) What is the heat of reaction for \(P=1 \mathrm{~atm}\) and \(T=894 \mathrm{~K}\) ?

For a low-density gas the virial expansion can be terminated at first order in the density and the equation of state is \( P=\frac{N k_{\mathrm{B}} T}{V}\left[1+\frac{N}{V} B_{2}(T)\right] $$ where \)B_{2}(T)\( is the second virial coefficient. The heat capacity will have corrections to its ideal gas value. We can write it in the form $$ C_{V, N}=\frac{3}{2} N k_{\mathrm{B}}-\frac{N^{2} k_{\mathrm{B}}}{V} F(T) $$ (a) Find the form that \)F(T)\( must have in order for the two equations to be thermodynamically consistent. (b) Find \)C_{P, N}$. (c) Find the entropy and internal energy.

A monatomic fluid in equilibrium is contained in a large insulated box of volume \(V\). The fluid is divided (conceptually) into \(m\) cells, each of which has an average number of particles \(N_{0}\), where \(N_{0}\) is large (neglect coupling between cells). Compute the variance in fluctuations of internal energy per particle \(u=U / N,\left\langle\left(\Delta u_{i}\right)^{2}\right\rangle\), in the ith cell. (Hint: Use temperature \(T\) and volume per particle \(v=V / N\) as independent variables.)

Compute the entropy, enthalpy, Helmholtz free energy, and Gibbs free energy of a paramagnetic substance and write them explicitly in terms of their natural variables when possible. Assume that the mechanical equation of state is \(m=(D H / T)\) and that the molar heat capacity at constant magnetization is \(c_{\mathrm{m}}=c\), where \(m\) is the molar magnetization, \(H\) is the magnetic field, \(D\) is a constant, \(c\) is a constant, and \(T\) is the temperature.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free