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Chapter 20: Problem 59

On a smoggy day in a certain city the ozone concentration was 0.42 ppm by volume. Calculate the partial pressure of ozone (in atm) and the number of ozone molecules per liter of air if the temperature and pressure were \(20.0^{\circ} \mathrm{C}\) and \(748 \mathrm{mmHg}\) respectively.

Short Answer

Expert verified
The partial pressure of the ozone is \(4.13 * 10^{-7} atm\). The number of ozone molecules per liter of air is \(1.03 * 10^{13}\) molecules per liter.

Step by step solution

01

Convert the pressure from mmHg to atm

Pressure, \(P=748mmHg\). To convert this to atmospheres (atm), use the conversion factor \(1 atm = 760 mmHg\). That is, \(P = 748/760 = 0.9842 atm\).
02

Convert ozone concentration from ppm to mole fraction

The ozone concentration, \(C=0.42ppm\), is given in parts per million but it is easier to work in terms of mole fractions with fractions of 1 being equivalent to 1 mole of gas. To convert to a mole fraction, divide by \(10^6\). Therefore, the mole fraction of ozone is \(0.42/10^6 = 4.20*10^{-7}\).
03

Calculate partial pressure of ozone

The partial pressure of a component in a gas mixture is equal to the mole fraction of that component times the total pressure. Therefore, the partial pressure of ozone, \(P_{O_3}\), can be calculated by multiplying the total pressure by the mole fraction of ozone. It can be calculated as: \(P_{O_3} = 0.9842 atm * 4.20*10^{-7}= 4.13 *10^{-7} atm\).
04

Calculate number of ozone molecules per liter of air

To find the number of ozone molecules per liter, it is necessary to use another form of the ideal gas law: \( PV = nRT\), which can be rearranged to find the number of moles per litre, n/V: \( n/V = P/RT\), substituting \(P_{O_3}\) for P and using the ideal gas constant, \(R = 0.0821 L.atm/(K.mol)\), gives \(n/V = 4.13 *10^{-7} atm/ (0.0821 L.atm/K.mol * 293.15K) = 1.71*10^{-11} mol/L\). Since there are \(6.022*10^{23}\) molecules in a mole, the number of ozone molecules per liter of air is \(1.71*10^{-11} mol/L * 6.022*10^{23} molecules/mol = 1.03*10^{13} molecules/L\).

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Most popular questions from this chapter

In 1991 it was discovered that nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) is produced in the synthesis of nylon. This compound, which is released into the atmosphere, contributes both to the depletion of ozone in the stratosphere and to the greenhouse effect. (a) Write equations representing the reactions between \(\mathrm{N}_{2} \mathrm{O}\) and oxygen atoms in the stratosphere to produce nitric oxide (NO), which is then oxidized by ozone to form nitrogen dioxide. (b) Is \(\mathrm{N}_{2} \mathrm{O}\) a more effective greenhouse gas than carbon dioxide? Explain. (c) One of the intermediates in nylon manufacture is adipic acid [HOOC(CH \(\left._{2}\right)_{4}\) COOH]. About \(2.2 \times 10^{9} \mathrm{~kg}\) of adipic acid are consumed every year. It is estimated that for every mole of adipic acid produced, 1 mole of \(\mathrm{N}_{2} \mathrm{O}\) is generated. What is the maximum number of moles of \(\mathrm{O}_{3}\) that can be destroyed as a result of this process per year?

Identify the gas that is responsible for the brown color of photochemical smog.

A person was found dead of carbon monoxide poisoning in a well-insulated cabin. Investigation showed that he had used a blackened bucket to heat water on a butane burner. The burner was found to function properly with no leakage. Explain, with an appropriate equation, the cause of his death.

The effective incoming solar radiation per unit area on Earth is \(342 \mathrm{~W} / \mathrm{m}^{2}\). Of this radiation, \(6.7 \mathrm{~W} / \mathrm{m}^{2}\) is absorbed by \(\mathrm{CO}_{2}\) at \(14,993 \mathrm{nm}\) in the atmosphere. How many photons at this wavelength are absorbed per second in \(1 \mathrm{~m}^{2}\) by \(\mathrm{CO}_{2} ?(1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s})\)

Instead of monitoring carbon dioxide, suggest another gas that scientists could study to substantiate the fact that \(\mathrm{CO}_{2}\) concentration is steadily increasing in the atmosphere.
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