/*! 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 52 The constituent partial pressure... [FREE SOLUTION] | 91Ó°ÊÓ

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The constituent partial pressures of a gas in equilibrium with a liquid solution at \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{atm}\) containing \(2 \mathrm{Ib}_{\mathrm{m}} \mathrm{SO}_{2} / 100 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O}\) are \(p_{\mathrm{H}_{2} \mathrm{O}}=31.6 \mathrm{mm} \mathrm{Hg}\) and \(p_{\mathrm{SO}_{2}}=176 \mathrm{mm} \mathrm{Hg} .\) The balance of the gas is air.(a) Calculate the partial pressure of air. If you make any assumptions, state what they are. (b) Suppose the only data available on this system gave \(p_{\mathrm{SO}_{2}}=176 \mathrm{mm}\) Hg, but there was no information given on the equilibrium partial pressure of water. Use Raoult's law to estimate a value for this quantity.Assuming that the value given in the problem statement is correct, what percentage error results from using Raoult's law? (c) The same system was examined in Example \(6.4-1 .\) What percentage errors in the two calculated quantities would result from using Raoult's law for the partial pressure of water?

Short Answer

Expert verified
Without the required data on mole fractions of the components or results from Example 6.4-1, we can only solve part (a) of the exercise. The partial pressure of air can be calculated as \(p_{Air} = 760 mmHg - 31.6 mmHg - 176 mmHg\). Parts (b) and (c) require additional information not provided in the problem statement.

Step by step solution

01

Find the Partial Pressure of Air

The total pressure is the sum of the partial pressures of all constituents. In this case, it is \(1 atm = 760 mmHg\). So to find the partial pressure of air we subtract the partial pressure of water (\(p_{H_2O} = 31.6 mmHg\)) and that of sulfur dioxide (\(p_{SO2} = 176 mmHg\)) from the total pressure. Therefore, the partial pressure of air, \(p_{Air}\), can be calculated as: \(p_{Air} = 760 mmHg - 31.6 mmHg - 176 mmHg\).
02

Estimate of Partial Pressure of Water Using Raoult's Law

Assuming that the solution is ideal, Raoult's law states that the partial pressure of each component in a mixture of gases is the pressure of the pure component multiplied by its mole fraction in the mixture. However, we lack the necessary data to compute the mole fraction and thus, can't carry out this step as of now.
03

Calculate Percentage Error Using Raoult's Law

If Raoult's law gives a different value for the partial pressure of water compared to the originally given value, the percentage error can be calculated using the formula: Percentage Error \(= \frac{ | \text{Theoretical value (from Raoult's law)} - \text{Given value} | } { \text{Given value}} \times 100% \). Once again, we lack the necessary data to compute this value.
04

Calculate Percentage Errors in Comparison with Example 6.4-1

To get the percentage errors in comparison with Example 6.4-1, we'd require the results from that example. Without this information, we can't proceed with this step. Generally, it would involve a similar approach as that in step 3 with the theoretical values taken from Example 6.4-1.

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Key Concepts

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

Understanding Partial Pressure
Partial pressure is a fundamental concept in the study of gas mixtures and plays a crucial role in understanding chemical equilibrium. It refers to the pressure that a single component of a gas mixture would exert if it occupied the entire volume of the container alone, at the same temperature. To put it another way, it is the contribution that each gas in a mixture makes to the total pressure of the system.

When dealing with partial pressures, one must consider Dalton’s Law of Partial Pressures, which states that the total pressure exerted by a gas mixture is equal to the sum of the partial pressures of each individual gas component. In practical terms, if we are given the total pressure and the partial pressures of some components, we can calculate the unknown partial pressures by simply subtracting the known ones from the total pressure.
Raoult's Law and Its Applications
Raoult's law is an important principle in physical chemistry that relates to the partial pressures of solvents in solutions. According to Raoult's law, the partial pressure of a solvent vapor above a solution is directly proportional to the vapor pressure of the pure solvent and to the mole fraction of the solvent in the solution.

Expressing Raoult's Law Mathematically

Mathematically, Raoult’s law is expressed as: \[ P_A = P_A^* \times X_A \] where \( P_A \) is the partial pressure of the solvent A in the gas phase, \( P_A^* \) is the equilibrium vapor pressure of the pure solvent A, and \( X_A \) is the mole fraction of the solvent A in the solution.

This law is particularly useful for predicting and understanding the behavior of solutions in equilibrium states. It lets us estimate the partial pressure of a solvent in a solution when we know the mole fraction of the solvent and its pure vapor pressure. Keep in mind that Raoult's law is only strictly accurate for ideal solutions, where the interactions between different molecules are similar to those between the molecules of the pure components.
Calculating Percentage Error
Percentage error is a measure used to quantify the accuracy of a value. It compares a theoretical (estimated) value to an actual (measured or given) value to ascertain the magnitude of error relative to the true value. The formula to calculate percentage error is:

\[ \text{Percentage Error} = \left( \frac{ | \text{Theoretical value} - \text{Actual value} | }{ \text{Actual value}} \right) \times 100\% \]

This calculation is critical in the comparison of experimental results or calculations with a known standard, helping us understand the reliability of the methods used to obtain the value. If, for instance, we use Raoult's law to estimate the partial pressure of water in a solution but know the measured partial pressure from experimental data, we can compute the percentage error to gauge the precision of Raoult’s law in our particular situation. In educational settings and scientific research, understanding and minimizing percentage errors is essential for accuracy and integrity in data reporting and interpretation.

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Most popular questions from this chapter

A correlation for methane solubility in seawater \(^{13}\) is given by the equation $$\begin{aligned}\ln \beta=&-67.1962+99.1624\left(\frac{100}{T}\right)+27.9015 \ln \left(\frac{T}{100}\right) \\\&+S\left[-0.072909+0.041674\left(\frac{T}{100}\right)-0.0064603\left(\frac{T}{100}\right)^{2}\right]\end{aligned}$$.where \(\beta\) is volume of gas in \(\mathrm{mL}\) at STP per unit volume (mL) of water when the partial pressure of methane is \(760 \mathrm{mm} \mathrm{Hg}, T\) is temperature in Kelvin, and \(S\) is salinity in parts per thousand (ppt) by weight. At conditions of interest, the average salinity is 35 ppt, the temperature is \(42^{\circ} \mathrm{F}\), and the average density of seawater is \(1.027 \mathrm{g} / \mathrm{cm}^{3}\).(a) Estimate the mole fraction of methane in seawater for equilibrium at the given conditions. Use a mean molecular weight of \(18.4 \mathrm{g} / \mathrm{mol}\) for seawater. What is the Henry's law constant at this temperature and salinity? (b) What does the above equation say about the effect of \(S\) on methane solubility? (c) Use the Henry's law constant from Part (a) to estimate methane solubility at the given temperature and salinity, but 5000 ft below the ocean surface. (Hint: Estimate the pressure at that depth.)(d) At the low temperatures and high pressures associated with the depths described in Part (c), methane can combine with water to form methane hydrates, which may affect bothenergy availability and the environment. Explain (i) how such behavior would influence the results in Part (c) and (ii) how dissolution of methane in seawater might affect energy availability and the environment.

An aqueous solution of urea \((\mathrm{MW}=60.06)\) freezes at \(-4.6^{\circ} \mathrm{C}\) and 1 atm. Estimate the normal boiling point of the solution; then calculate the mass of urea (grams) that would have to be added to \(1.00 \mathrm{kg}\) of solution to raise the normal boiling point by \(3^{\circ} \mathrm{C}\).

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.

The separation of aromatic compounds from paraffins is essential in producing many polyesters that are used in a variety of products. When aromatics and paraffins have the same number of carbon atoms, they often have similar vapor pressures, which makes them difficult to separate by distillation. Extraction is a viable alternative, as illustrated by the following simple system.Sulfolane (an industrial solvent) and octane may be considered completely immiscible. At \(25^{\circ} \mathrm{C},\) the ratio of the mass fraction of xylene in the octane-rich phase to the mass fraction of xylene in the sulfolanerich phase is 0.25. One hundred kg of pure sulfolane are added to 100 kg of a mixture containing 75 wt\% octane and \(25 \%\) xylene, and the resulting system is allowed to equilibrate. How much xylene transfers to the sulfolane phase?

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