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The temperature inside a pressure cooker is \(115^{\circ} \mathrm{C}\). Calculate the vapor pressure of water inside the pressure cooker. What would be the temperature inside the pressure cooker if the vapor pressure of water was \(3.50 \mathrm{~atm}\) ?

Short Answer

Expert verified
The vapor pressure of water inside the pressure cooker at \(115^\circ C\) is approximately 2.33 atm. When the vapor pressure of water is 3.50 atm, the temperature inside the pressure cooker is approximately 137.84°C.

Step by step solution

01

Understand the Antoine equation

The Antoine equation is a function that provides the vapor pressure of a substance, given its temperature. The equation is in the form: \[P(T) = 10^{A - \frac{B}{T+C}}\] Where: - \(P(T)\): Vapor pressure of the substance at temperature T - T: Temperature in Celsius - A, B, and C: Antoine coefficients; these are substance-specific constants For water, values for A, B, and C can be found from textbooks or online resources. In this exercise, we will use the following values: A = 8.07131 B = 1730.63 C = 233.426
02

Calculate the vapor pressure of water

We are given the temperature (\(T = 115^\circ C\)), so we can plug this value into the Antoine equation with the given coefficients to calculate the vapor pressure. \[P(115) = 10^{8.07131 - \frac{1730.63}{115+233.426}}\] Using a calculator, we find that the vapor pressure of water inside the pressure cooker is approximately 2.33 atm.
03

Calculate the temperature with the given vapor pressure

We are given a vapor pressure (\(P = 3.50 \,atm\)), and we want to find the temperature. This time, we will rearrange the Antoine equation to solve for T. \[P(T) = 10^{A - \frac{B}{T+C}}\] \[3.50 = 10^{8.07131 - \frac{1730.63}{T+233.426}}\] We will first take the log base 10 of both sides: \[log_{10}(3.50) = 8.07131 - \frac{1730.63}{T+233.426}\] Next, we will rearrange the equation to derive a value for T: \[T = \frac{B}{A - log_{10}(3.50)} - C = \frac{1730.63}{8.07131 - log_{10}(3.50)} - 233.426\] Using a calculator, we find that the temperature inside the pressure cooker when the vapor pressure of water is 3.50 atm is approximately 137.84°C.

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Key Concepts

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

Antoine Equation
The Antoine equation is an essential tool in physical chemistry for estimating the vapor pressure of a substance at a specific temperature. It's prevalently used because of its simplicity and reliable accuracy for many applications, such as in distillation processes. In the realm of thermodynamics, understanding vapor pressure is crucial as it reflects how a substance behaves when transitioning between liquid and gaseous states.

The equation itself has a logarithmic form and is expressed as:\[P(T) = 10^{A - \frac{B}{T+C}}\]In this relationship, the symbols represent the following:
  • \(P(T)\): Vapor pressure at temperature \(T\).
  • \(T\): Temperature in Celsius at which the pressure is being evaluated.
  • \(A, B, C\): Substance-specific constants known as Antoine coefficients.
The Antoine equation accommodates the non-linear relationship between temperature and vapor pressure. To use it, you'll need the specific Antoine coefficients for the substance in question. For water, these coefficients are typically provided in chemistry references and must be used at temperature ranges for which they were determined to ensure accuracy.

Detailed explanations, such as recognizing that different substances have unique sets of coefficients, aid in comprehending the equation's precise functionality. Moreover, clarifying that these coefficients can alter based on the pressure unit used reinforces its practical understanding.
Vapor Pressure of Water
When discussing the vapor pressure of water, one refers to the pressure exerted by water vapor when water is in equilibrium with its liquid or solid form. At this point, the rate of water molecules evaporating from the liquid (or sublimating from the solid) equals the rate of water molecules condensing back into the liquid or solid phase. The vapor pressure is a dynamic attribute, sensitive to changes in temperature: as temperature goes up, so does the vapor pressure.

Understanding vapor pressure is fundamental in a wide array of fields, including meteorology, where it contributes to weather predictions, and engineering, particularly in designs involving boilers or condensers. Additionally, vapor pressure influences various everyday occurrences such as the boiling of water, humidity inside homes, and even the preservation of foods in sealed containers.

The vapor pressure can also signal a substance's volatility—the higher the vapor pressure at a given temperature, the more volatile the substance. Moreover, by elaborating that the vapor pressure of water at normal boiling point is equivalent to atmospheric pressure, the concept bridges into practical scenarios, further aiding student comprehension.
Pressure Cooker Temperature
A pressure cooker takes advantage of the relationship between pressure and the boiling point of water to cook food more swiftly. By sealing the container, a pressure cooker allows pressure to build as water is heated, which in turn elevates the boiling point of the water inside. Higher boiling points mean that food can be cooked at higher temperatures, thus shortening cooking times drastically.

The increased pressure within the pressure cooker has a direct effect on the vapor pressure—this pressure corresponds to the vapor pressure of water at the elevated boiling point. Hence, when the temperature rises within a pressure cooker, so does the internal vapor pressure, until it surpasses normal atmospheric pressure resulting in higher cooking temperatures. For instance, at sea level, water boils at 100°C at 1 atmospheric pressure, whereas inside a pressure cooker, temperatures can exceed 120°C.

By incorporating the element of safety relating to handling high-pressure devices and explaining that the cooker's design allows for this pressure build-up without the risk of immediate evaporation or boiling over, the practical understanding is significantly amplified. This not only helps to appreciate the cooking technology but also solidifies the scientific principles governing phase changes in response to temperature and pressure variations.

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Most popular questions from this chapter

An ice cube tray contains enough water at \(22.0^{\circ} \mathrm{C}\) to make 18 ice cubes that each has a mass of \(30.0 \mathrm{~g}\). The tray is placed in a freezer that uses \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) as a refrigerant. The heat of vaporization of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) is \(158 \mathrm{~J} / \mathrm{g}\). What mass of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) must be vaporized in the refrigeration cycle to convert all the water at \(22.0^{\circ} \mathrm{C}\) to ice at \(-5.0^{\circ} \mathrm{C}\) ? The heat capacities for \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are \(2.03 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the enthalpy of fusion for ice is \(6.02 \mathrm{~kJ} / \mathrm{mol}\).

Carbon diselenide \(\left(\mathrm{CSe}_{2}\right)\) is a liquid at room temperature. The normal boiling point is \(125^{\circ} \mathrm{C}\), and the melting point is \(-45.5^{\circ} \mathrm{C}\). Carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) is also a liquid at room temperature with normal boiling and melting points of \(46.5^{\circ} \mathrm{C}\) and \(-111.6^{\circ} \mathrm{C}\), respectively. How do the strengths of the intermolecular forces vary from \(\mathrm{CO}_{2}\) to \(\mathrm{CS}_{2}\) to \(\mathrm{CSe}_{2}\) ? Explain.

General Zod has sold Lex Luthor what Zod claims to be a new copper-colored form of kryptonite, the only substance that can harm Superman. Lex, not believing in honor among thieves, decided to carry out some tests on the supposed kryptonite. From previous tests, Lex knew that kryptonite is a metal having a specific heat capacity of \(0.082 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and a density of \(9.2 \mathrm{~g} / \mathrm{cm}^{3}\) Lex Luthor's first experiment was an attempt to find the specific heat capacity of kryptonite. He dropped a \(10 \mathrm{~g} \pm 3 \mathrm{~g}\) sample of the metal into a boiling water bath at a temperature of \(100.0^{\circ} \mathrm{C} \pm 0.2^{\circ} \mathrm{C}\). He waited until the metal had reached the bath temperature and then quickly transferred it to \(100 \mathrm{~g} \pm\) \(3 \mathrm{~g}\) of water that was contained in a calorimeter at an initial temperature of \(25.0^{\circ} \mathrm{C} \pm 0.2^{\circ} \mathrm{C}\). The final temperature of the metal and water was \(25.2^{\circ} \mathrm{C}\). Based on these results, is it possible to distinguish between copper and kryptonite? Explain. When Lex found that his results from the first experiment were inconclusive, he decided to determine the density of the sample. He managed to steal a better balance and determined the mass of another portion of the purported kryptonite to be \(4 \mathrm{~g} \pm 1 \mathrm{~g}\). He dropped this sample into water contained in a \(25-\mathrm{mL}\) graduated cylinder and found that it displaced a volume of \(0.42 \mathrm{~mL} \pm 0.02 \mathrm{~mL}\). Is the metal copper or kryptonite? Explain. Lex was finally forced to determine the crystal structure of the metal General Zod had given him. He found that the cubic unit cell contained four atoms and had an edge length of 600\. pm. Explain how this information enabled Lex to identify the metal as copper or kryptonite. Will Lex be going after Superman with the kryptonite or seeking revenge on General Zod? What improvements could he have made in his experimental techniques to avoid performing the crystal structure determination?

The critical point of \(\mathrm{NH}_{3}\) is \(132^{\circ} \mathrm{C}\) and \(111 \mathrm{~atm}\), and the critical point of \(\mathrm{N}_{2}\) is \(-147^{\circ} \mathrm{C}\) and \(34 \mathrm{~atm}\). Which of these substances cannot be liquefied at room temperature no matter how much pressure is applied? Explain.

A \(20.0-\mathrm{g}\) sample of ice at \(-10.0^{\circ} \mathrm{C}\) is mixed with \(100.0 \mathrm{~g}\) water at \(80.0^{\circ} \mathrm{C}\). Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are \(2.03\) and \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the enthalpy of fusion for ice is \(6.02 \mathrm{~kJ} / \mathrm{mol}\).

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