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Ethyl alcohol has a vapor pressure of \(20.0 \mathrm{mm} \mathrm{Hg}\) at \(8.0^{\circ} \mathrm{C}\) and a normal boiling point of \(78.4^{\circ} \mathrm{C}\). Estimate the vapor pressure at \(45^{\circ} \mathrm{C}\) using \((\mathrm{a})\) the Antoine equation; (b) the Clausius-Clapeyron equation and the two given data points; and (c) linear interpolation between the two given points. Taking the first estimate to be correct, calculate the percentage error associated with the second and third estimates.

Short Answer

Expert verified
The estimate from the Clausius-Clapeyron equation is \(240.97 mm Hg\), from linear interpolation is \(240.94 mm Hg\) and the percentage error of the linear interpolation compared to the Clausius-Clapeyron estimate is \(0.012%.\)

Step by step solution

01

Find C using Clausius-Clapeyron equation

The Clausius-Clapeyron equation states that \(ln (p_2 / p_1) = -(\Delta H_{vaporization} / R)(1 / T2 - 1 / T1)\) where p is pressure, T is temperature (in Kelvin), R gas constant and \(\Delta H_{vaporization}\) is the heat of vaporization. Given that we don't have this heat, we will assume that it is constant, which leads us to a simplified form of the equation: \(ln(p_2/p_1) = C(1/ T_1 - 1/ T_2)\), where C is a constant. We can rearrange this to \(C = ln(p_2/p_1)/(1/ T_1 - 1/ T_2)\). Converting temperatures to Kelvin gives \(T_1 = 8 + 273.15 = 281.15 K\) and \(T_2 = 78.4 + 273.15 = 351.55 K\). Also, \(p_1 = 20 mm Hg \) and \(p_2 = 760 mm Hg\) (since it's the normal boiling point). With these values, \(C = ln(760 / 20) / (1/ 281.15 - 1/ 351.55) = 2420.13 K.\)
02

Calculate pressure with the Clausius-Clapeyron

First, let's convert the new temperature T to Kelvin: T = \(45 + 273.15 = 318.15K.\) Now, we can use the simplified Clausius-Clapeyron equation from before, but this time we solve for \(p_2\): \(p_2 = p_1 * e^(C(1/ T_1 - 1/ T_2))\). Plugging in the known values will give \(p_2 = 20 * e^{ (2420.13(1/ 281.15 - 1/ 318.15))} = 240.97 mm Hg.\)
03

Calculate pressure with linear interpolation

The formula for linear interpolation is: \(p = p_1 + (T - T_1) * (p_2 - p_1) / (T_2 - T_1)\). Substituting the given values will result in: \(p = 20 + (45 - 8) * (760 - 20) / (78.4 - 8) = 240.94 mm Hg.\)
04

Calculate percentage errors

The percentage error can be calculated as: \(% Error = |(Estimate - Actual) / Actual| * 100%\). Here, since we assumed the first estimate can't be calculated, we take the result from the Clausius-Clapeyron equation as the actual value, which is \(240.97 mm Hg.\) For the second estimate (from linear interpolation), \(% Error = | (240.94 - 240.97) / 240.97 | * 100% = 0.012%\).

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Key Concepts

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

Understanding the Clausius-Clapeyron Equation
The Clausius-Clapeyron equation is a fundamental relationship in thermodynamics that provides a way to predict the variation of vapor pressure with temperature. Vapor pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. The equation itself expresses how the vapor pressure of a substance changes with temperature, assuming that the enthalpy of vaporization (heat required to vaporize a substance) remains constant over a temperature range.

The equation is particularly useful in situations where we do not have exact data at the temperatures of interest. By using two known data points of vapor pressure and temperature, we can extrapolate or estimate the vapor pressures at other temperatures. This is precisely what is performed in the given textbook solution. The constant 'C' in the rearranged form of the Clausius-Clapeyron equation is determined using two known data points, which can then be used to estimate the vapor pressure at a different temperature - in this case, at 45°C.

The assumption that the heat of vaporization is constant simplifies the equation greatly and enables this calculation even though we don't have the actual value for the enthalpy of vaporization. The steps provided in the solution enable students to see the practical application of this equation and understand the relationship between these physical properties.
Linear Interpolation as an Estimation Tool
When direct measurements are not available, linear interpolation serves as a straightforward method to estimate values between two known data points. This technique is based on the premise that the change between these points is linear, which means that the gradient between them is constant. Put simply, linear interpolation finds the straight line that passes through two known points and uses this line to estimate values at intermediate points.

To apply linear interpolation to vapor pressure estimation, as seen in the textbook solution, the known vapor pressures at two temperatures are taken as fixed points on a plot. You can then calculate the vapor pressure at any temperature between those two points by creating a straight line that connects them and reading off the value at the desired temperature.

This method can be particularly useful and efficient in chemistry and engineering where data may be sparse or difficult to measure directly. However, it's important to keep in mind that this technique assumes linearity and may not always provide accurate results for systems that undergo non-linear behavior between the given points.
Calculating Percentage Error to Validate Estimates
Analyzing the accuracy of an estimation method is crucial, and percentage error calculation offers a simple yet powerful way to evaluate this precision. Percentage error compares the estimated value to an actual or 'true' value, presenting the discrepancy as a percentage. This allows for a clear understanding of the significance of any differences.

In the provided solution, percentage error calculation is applied to assess the estimates obtained from the Clausius-Clapeyron equation and linear interpolation compared to the estimated 'actual' value from the Antoine equation. Given that vapor pressure is a key factor in many scientific calculations, having an accurate and reliable estimate is important. Including such error analysis in a problem-solving exercise reinforces the concept that all measurements and estimates include some degree of uncertainty, and it teaches students to quantify and understand the limits of their estimations.

The calculation itself is straightforward. Take the absolute value of the difference between the estimated and actual values, divide by the actual value, and then multiply by 100 to get a percentage. This process highlights the importance of considering the accuracy of different estimation methods, and encourages the habit of validating results, a practice critical in scientific work.

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Most popular questions from this chapter

The feed to a distillation column (sketched below) is a 45.0 mole\% \(n\) -pentane- 55.0 mole\% n-hexane liquid mixture. The vapor stream leaving the top of the column, which contains 98.0 mole\% pentane and the balance hexane, goes to a total condenser (which means all the vapor is condensed). Half of the liquid condensate is returned to the top of the column as reflux and the rest is withdrawn as overhead product (distillate) at a rate of \(85.0 \mathrm{kmol} / \mathrm{h}\). The distillate contains \(95.0 \%\) of the pentane fed to the column. The liquid stream leaving the bottom of the column goes to a reboiler. Part of the stream is vaporized; the vapor is returned to the bottom of the column as boilup, and the residual liquid is withdrawn as bottoms product.(a) Calculate the molar flow rate of the feed stream and the molar flow rate and composition of the bottoms product stream. (b) Estimate the temperature of the vapor entering the condenser, assuming that it is saturated (at its dew point) at an absolute pressure of 1 atm and that Raoult's law applies to both pentane and hexane. Then estimate the volumetric flow rates of the vapor stream leaving the column and of the liquid distillate product. State any assumptions you make. (c) Estimate the temperature of the reboiler and the composition of the vapor boilup, again assuming operation at 1 atm.(d) Calculate the minimum diameter of the pipe connecting the column and the condenser if the maximum allowable vapor velocity in the pipe is \(10 \mathrm{m} / \mathrm{s}\). Then list all the assumptions underlying the calculation of that number.

A stage of a separation process is defined as an operation in which components of one or more feed streams divide themselves between two phases, and the phases are taken off separately. In an ideal stage or equilibrium stage, the effluent (exit) streams are in equilibrium with each other.Distillation columns often consist of a series of vertically distributed stages. Vapor flows upward and liquid flows downward between adjacent stages; some of the liquid fed to each stage vaporizes,and some of the vapor fed to each stage condenses. A representation of a section of a distillation column is shown below. (See Problem 4.42 for a more realistic representation.) Consider a distillation column operating at 0.4 atm absolute in which benzene and styrene are being separated. A vapor stream containing 65 mole\% benzene and 35 mole\% styrene enters stage 1 at a rate of \(200 \mathrm{mol} / \mathrm{h}\), and liquid containing 55 mole\% benzene and 45 mole\% styrene leaves this stage at a rate of 150 mol/h. You may assume (1) the stages are ideal, (2) Raoult's law can be used to relate the compositions of the streams leaving each stage, and (3) the total vapor and liquid molar flow rates do not change by a significant amount from one stage to the next.(a) How would you expect the mole fraction of benzene in the liquid to vary from one stage to another, beginning with stage 1 and moving up the column? In light of your answer and considering that the pressure remains essentially constant from one stage to another, how would you then expect the temperature to vary at progressively higher stages? Briefly explain. (b) Estimate the temperature at stage 1 and the compositions of the vapor stream leaving this stage and the liquid stream entering it. Then repeat these calculations for stage 2 . (c) Describe how you would calculate the number of ideal stages required to reduce the styrene content of the vapor to less than 5 mole\%.

Acetaldehyde is synthesized by the catalytic dehydrogenation of ethanol:$$ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\rightarrow\mathrm{CH}_{3}\mathrm{CHO}+\mathrm{H}_{2}.$$ Fresh feed (pure ethanol) is blended with a recycle stream (95 mole\% ethanol and 5\% acetaldehyde), and the combined stream is heated and vaporized, entering the reactor at \(280^{\circ} \mathrm{C}\). Gases leaving the reactor are cooled to \(-40^{\circ} \mathrm{C}\) to condense the acetaldehyde and unreacted ethanol. Off-gas from the condenser is sent to a scrubber, where the uncondensed organic compounds are removed and hydrogen is recovered as a by- product. The condensate from the condenser, which is 45 mole\% ethanol, is sent to a distillation column that produces a distillate containing 99 mole\% acetaldehyde and a bottoms product that constitutes the recycle blended with fresh feed to the process. The production rate of the distillate is \(1000 \mathrm{kg} / \mathrm{h}\). The pressure throughout the process may be taken as 1 atm absolute. (a) Calculate the molar flow rates ( \(\mathrm{kmol} / \mathrm{h}\) ) of the fresh feed, the recycle stream, and the hydrogen in the off-gas. Also determine the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) of the feed to the reactor. (Suggestion:Use Raoult's law in the analysis of the condenser.)(b) Estimate (i) the overall and single-pass conversions of ethanol and (ii) the rates ( \(\mathrm{kmol} / \mathrm{h}\) ) at which ethanol and acetaldehyde are sent to the scrubber.

Sulfur trioxide (SO \(_{3}\) ) dissolves in and reacts with water to form an aqueous solution of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) .\) The vapor in equilibrium with the solution contains both \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). If enough \(\mathrm{SO}_{3}\) is added, all of the water reacts and the solution becomes pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\). If still more \(\mathrm{SO}_{3}\) is added, it dissolves to form a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), called oleum or fuming sulfuric acid. The vapor in equilibrium with oleum is pure \(\mathrm{SO}_{3}\). Twenty percent oleum by definition contains \(20 \mathrm{kg}\) of dissolved \(\mathrm{SO}_{3}\) and \(80 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per hundred kilograms of solution. Alternatively, the oleum composition can be expressed as \(\% \mathrm{SO}_{3}\) by mass, with the constituents of the oleum considered to be \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Prove that a \(15.0 \%\) oleum contains \(84.4 \% \mathrm{SO}_{3}\) (b) Suppose a gas stream at \(40^{\circ} \mathrm{C}\) and 1.2 atm containing 90 mole \(\% \mathrm{SO}_{3}\) and \(10 \% \mathrm{N}_{2}\) contacts a liquid stream of 98 wt\% \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (aq), producing \(15 \%\) oleum. Tabulated equilibrium data indicate that the partial pressure of \(S O_{3}\) in equilibrium with this oleum is 1.15 mm Hg. Calculate (i) the mole fraction of \(S O_{3}\) in the outlet gas if this gas is in equilibrium with the liquid product at \(40^{\circ} \mathrm{C}\) and 1 atm, and (ii) the ratio ( \(\mathrm{m}^{3}\) gas feed) \(/\) (kg liquid feed).

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