/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Ethyl alcohol has a vapor pressu... [FREE SOLUTION] | 91Ó°ÊÓ

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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

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