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Chapter 10: Problem 72

(a) What are the mole fractions of each component in a mixture of $15.08 \mathrm{~g}\( of \)\mathrm{O}_{2}, 8.17 \mathrm{~g}\( of \)\mathrm{N}_{2},$ and \(2.64 \mathrm{~g}\) of \(\mathrm{H}_{2}\) (b) What is the partial pressure in atm of each component of this mixture if it is held in a \(15.50-\mathrm{L}\) vessel at \(15^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The mole fractions of O2, N2, and H2 in the mixture are 0.22624, 0.14008, and 0.63368, respectively. The partial pressures of O2, N2, and H2 are 0.55001 atm, 0.34075 atm, and 1.54161 atm, respectively.

Step by step solution

01

Find moles of each component

In order to find the moles of each gas, we will use their molar masses: Molar Mass of O2: 32 g/mol Molar Mass of N2: 28 g/mol Molar Mass of H2: 2 g/mol Now, we will find the moles of each gas: moles of O2 = \( \frac{15.08 g}{32 g/mol} = 0.47125 mol\) moles of N2 = \( \frac{8.17 g}{28 g/mol} = 0.29196 mol\) moles of H2 = \( \frac{2.64 g}{2 g/mol} = 1.32 mol\)
02

Calculate mole fractions of each component

Mole fractions of each component in the mixture can be found using the following formula: \(Mole\,Fraction\,(X_i) = \frac{moles\,of\,component\,i}{total\,moles\,of\,all\,components}\) Total moles of all components = moles of O2 + moles of N2 + moles of H2 = 0.47125 + 0.29196 + 1.32 = 2.08321 Mole Fraction of O2 (\(X_{O_2}\)) = \( \frac{0.47125}{2.08321} = 0.22624\) Mole Fraction of N2 (\(X_{N_2}\)) = \( \frac{0.29196}{2.08321} = 0.14008 \) Mole Fraction of H2 (\(X_{H_2}\)) = \( \frac{1.32}{2.08321} = 0.63368 \)
03

Find the total pressure of the mixture

We can find the total pressure of the mixture using the ideal gas law: \(PV = nRT\) where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant (0.0821 L⋅atm/mol⋅K), and T is the temperature in Kelvin. First, we need to convert the temperature from degrees Celsius to Kelvin: Temperature in Kelvin = 15 + 273.15 = 288.15 K Now we can find the total pressure of the mixture. We will use the total moles from Step 2 and the given volume of the vessel (15.50 L): Total Pressure (P) = \( \frac{(2.08321\,mol)(0.0821\,\frac{L\cdot atm}{mol\cdot K})(288.15\,K)}{15.5\,L} = 2.43237\, atm\)
04

Calculate partial pressures of each component

To find the partial pressure of each component, we will use the mole fraction and the total pressure: Partial Pressure (\(P_i\)) = Mole Fraction (\(X_i\)) × Total Pressure (P) Partial Pressure of O2 (\(P_{O_2}\)) = 0.22624 × 2.43237 atm = 0.55001 atm Partial Pressure of N2 (\(P_{N_2}\)) = 0.14008 × 2.43237 atm = 0.34075 atm Partial Pressure of H2 (\(P_{H_2}\)) = 0.63368 × 2.43237 atm = 1.54161 atm The partial pressures of O2, N2, and H2 are 0.55001 atm, 0.34075 atm, and 1.54161 atm, respectively.

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Most popular questions from this chapter

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? (a) Lifting up on the piston to double the volume while keeping the temperature constant; (b) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\), while keeping the volume constant; (c) Pushing down on the piston to halve the volume while keeping the temperature constant.

Consider a mixture of two gases, \(\mathrm{A}\) and \(\mathrm{B}\), confined in a closed vessel. A quantity of a third gas, \(\mathrm{C}\), is added to the same vessel at the same temperature. How does the addition of gas \(\mathrm{C}\) affect the following: (a) the partial pressure of gas \(A,(\mathbf{b})\) the total pressure in the vessel, (c) the mole fraction of gas B?

At constant pressure, the mean free path \((\lambda)\) of a gas molecule is directly proportional to temperature. At constant temperature, \(\lambda\) is inversely proportional to pressure. If you compare two different gas molecules at the same temperature and pressure, \(\lambda\) is inversely proportional to the square of the diameter of the gas molecules. Put these facts together to create a formula for the mean free path of a gas molecule with a proportionality constant (call it \(R_{\mathrm{mfp}}\), like the ideal-gas constant) and define units for \(R_{\mathrm{mfp}^{-}}\)

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Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at $80^{\circ} \mathrm{C}\( if 1.00 mol occupies \)33.3 \mathrm{~L}$, assuming that (a) \(\mathrm{CCl}_{4}\) obeys the ideal-gas equation; (b) \(\mathrm{CCl}_{4}\) obeys the van der Waals equation. (Values for the van der Waals constants are given in Table \(10.3 .\) ) (c) Which would you expect to deviate more from ideal behavior under these conditions, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4} ?\) Explain.
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