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At a 4000 -m altitude the atmospheric pressure is about \(0.605 \mathrm{~atm}\). What boiling point would you expect for water under these conditions?

Short Answer

Expert verified
The boiling point of water at a 4000-m altitude with an atmospheric pressure of 0.605 atm is approximately \(78.75^\circ \text{C}\).

Step by step solution

01

Recall the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is given by: \[\ln \frac{P_2}{P_1} = -\frac{\Delta H_\text{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] where \(P_1\) and \(P_2\) are the vapor pressures at the temperatures \(T_1\) and \(T_2\) respectively, \(\Delta H_\text{vap}\) is the enthalpy of vaporization, and R is the ideal gas constant (8.314 J/mol·K).
02

Use the given pressure and standard boiling point information

We know that the atmospheric pressure at the 4000-m altitude is 0.605 atm, which is approximately 61221 Pa. We also have the standard atmospheric pressure as 101325 Pa, and the boiling point of water at standard atmospheric pressure as 373.15 K (100°C). So, we have: \(P_1 =\) 101325 Pa, \(T_1 =\) 373.15 K \(P_2 =\) 61221 Pa, \(T_2 =\) unknown
03

Determine the Enthalpy of Vaporization of Water

In order to use the Clausius-Clapeyron equation, we need to determine the enthalpy of vaporization of water. For water, the enthalpy of vaporization is approximately 40.7 kJ/mol, which can be converted to J/mol: \(\Delta H_\text{vap} =\) 40,700 J/mol
04

Solve for T2 using the Clausius-Clapeyron Equation

We can now plug in the given values and solve for \(T_2\): \[\ln \frac{61221}{101325} = -\frac{40,700}{8.314} \left(\frac{1}{T_2} - \frac{1}{373.15}\right)\] Solving for \(T_2\): \[T_2 = \frac{1}{\frac{1}{373.15} - \frac{8.314}{40,700} \ln\frac{61221}{101325}}\]
05

Calculate the boiling point of water at 4000-m altitude

By plugging the given values into the equation from Step 4, we get the value of \(T_2\): \[T_2 = \frac{1}{\frac{1}{373.15} - \frac{8.314}{40,700} \ln\frac{61221}{101325}} = 351.9 \text{ K}\] Now, we need to convert this temperature into degrees Celsius: Boiling Point \(= T_2 - 273.15 = 351.9 - 273.15 = 78.75^\circ \text{C}\) Thus, the boiling point of water under 0.605 atm pressure at a 4000-m altitude is approximately \(78.75^\circ \text{C}\).

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Key Concepts

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

Clausius-Clapeyron equation
The Clausius-Clapeyron equation is a vital tool in thermodynamics for analyzing the relationship between pressure and temperature concerning phase changes, such as from liquid to vapor.
This equation provides a way to calculate the boiling point of a substance at a different pressure.
The formula is:
  • \[\ln \frac{P_2}{P_1} = -\frac{\Delta H_\text{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]where:
  • \(P_1\) and \(P_2\) are the vapor pressures at the initial and final temperatures \(T_1\) and \(T_2\) respectively.
  • \(\Delta H_\text{vap}\) is the enthalpy of vaporization.
  • \(R\) is the ideal gas constant \((8.314 \text{ J/mol·K})\).
By using known pressures and temperatures, you can predict the temperature at which a substance will boil when the pressure changes. This is particularly useful for understanding how different environmental conditions affect boiling points.
This equation assumes constant enthalpy of vaporization over the temperature range, which is a good approximation for small differences in temperature.
enthalpy of vaporization
The enthalpy of vaporization is the amount of energy required to convert a liquid into a vapor at constant pressure. It reflects the energy needed to overcome intermolecular forces during the phase transition from liquid to gas.
For example, water has a relatively high enthalpy of vaporization, approximately 40.7 kJ/mol, which means significant energy is required to break the hydrogen bonds between water molecules.

Enthalpy of vaporization varies with temperature, but for applications in the Clausius-Clapeyron equation, it is often treated as a constant to simplify calculations.
  • A high enthalpy of vaporization means a substance needs more heat to evaporate.
  • It is important for understanding processes such as boiling, evaporation, and condensation.
Recognizing the enthalpy of vaporization allows scientists and engineers to predict and control temperature and pressure conditions for industrial applications and understand natural processes like the climate impact and weather patterns.
vapor pressure
Vapor pressure is the pressure exerted by the vapor of a liquid in equilibrium with its liquid phase at a given temperature. It's an important concept because it signifies the tendency of molecules to escape from the liquid phase to the vapor phase.
Higher temperature usually increases vapor pressure because molecules acquire more energy to break free from intermolecular attractions.

In practical terms, vapor pressure helps determine when a liquid will boil because boiling occurs when the vapor pressure equals the surrounding atmospheric pressure.
  • At higher vapor pressures, the molecules are more likely to transition from liquid to gas.
  • A substance with high vapor pressure at a specific temperature is considered volatile.
Understanding vapor pressure is crucial in various applications, including cooking, manufacturing, and meteorology, as it affects how substances interact with their environment and how pressure changes can alter boiling points.
altitude effects on boiling point
Altitude has a direct impact on the boiling point of water due to changes in atmospheric pressure. As altitude increases, atmospheric pressure decreases because there is less air above exerting pressure.
This drop in pressure reduces the boiling point of water, meaning water will boil at lower temperatures at higher altitudes.

For example, at sea level, water boils at 100°C, but at a 4000-meter altitude, where the pressure is around 0.605 atm, the boiling point lowers to approximately 78.75°C, as calculated in our example.
  • Lower boiling points can affect cooking times and the efficiency of heating processes.
  • It is important for pilots and climbers to be aware of these changes in boiling points for safety and operational purposes.
By understanding how altitude affects the boiling point, adjustments can be made to processes and recipes, ensuring safety and efficiency in high-altitude conditions.

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Most popular questions from this chapter

If the vapor pressure of \(\mathrm{CC} 1_{4}\) (carbon tetrachloride) is \(.132\) atm at \(23^{\circ} \mathrm{C}\) and \(.526 \mathrm{~atm}\) at \(58^{\circ} \mathrm{C}\), what is the \(\Delta \mathrm{H}^{\prime}\) in this temperature range?

The vapor pressure of benzene at \(75^{\circ} \mathrm{C}\) is 640 torr. \(\mathrm{A}\) solution of \(3.68 \mathrm{~g}\) of a solute in \(53.0 \mathrm{~g}\) benzene has a vapor pressure of 615 torr. Calculate the molecular weight of the solute. (MW of benzene \(=78.0 .)\)

A G.T.O. has a 22 gal. cooling system. Suppose you fill it with a \(50-50\) solution by volume of \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\), ethylene glycol, and water. At what temperature would freezing become a problem? Assume the specific gravity of ethylene glycol is \(1.115\) and the freezing point depression constant of water is \(\left(1.86 \mathrm{C}^{\circ} / \mathrm{mole}\right)\). You might have placed in methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) instead of \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\). If the current cost of ethylene glycol is 12 cents \(/ \mathrm{lb}\) and the cost of methanol is 8 cents \(/ \mathrm{lb}\), how much money would you save by using \(\mathrm{CH}_{3} \mathrm{OH}\) ? And yet, ethylene glycol is the more desirable antifreeze. Why? density \(=(.79 \mathrm{~g} / \mathrm{ml})\) for \(\mathrm{CH}_{3} \mathrm{OH}\) and \(3.785\) liter \(=1\) gallon.

A chemist wishes to determine the molecular weight and molecular formula of fructose (a sugar). He places \(.946 \mathrm{~g}\) of it in \(150 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (water) and finds that the freezing point of water is depressed to \(-0.0651^{\circ} \mathrm{C} .\) Determine the molecular weight and formula of fructose, assuming that the simplest formula of fructose is \(\left(\mathrm{CH}_{2}\right) \mathrm{O}\).

A solution of \(20.0 \mathrm{~g}\) of a non-volatile solute in \(100 \mathrm{~g}\) of benzene at \(30^{\circ} \mathrm{C}\) has a vapor pressure \(13.4\) torr lower than the vapor pressure of pure benzene. What is the mole fraction of solute? Vapor pressure of benzene at \(30^{\circ} \mathrm{C}=\) \(121.8\) torr.

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