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Chapter 10: Problem 64

A deep-sea diver uses a gas cylinder with a volume of \(10.0 \mathrm{~L}\) and a content of \(51.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(32.6 \mathrm{~g}\) of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is \(19^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The partial pressure of oxygen is \(391.54 \text{ Pa}\), the partial pressure of helium is \(1987.18 \text{ Pa}\), and the total pressure is \(2378.72 \text{ Pa}\).

Step by step solution

01

Convert the mass of each gas to moles

We need to determine the amount of each gas in moles. To do this, we will use the molar mass of each gas: Molar masses: Oxygen (O2): 32 g/mol Helium (He): 4 g/mol Number of moles (n) = mass (m) / molar mass Moles of O2: \(n_{O_2} = \frac{51.2 \text{ g}}{32 \text{ g/mol}} = 1.6 \text{ mol}\) Moles of He: \(n_{He} = \frac{32.6 \text{ g}}{4 \text{ g/mol}} = 8.15 \text{ mol}\)
02

Convert temperature to Kelvin

To use the Ideal Gas Law, we need to convert the Celsius temperature given into Kelvin. \(T (K) = T (^\circ C) + 273.15\) \(T = 19^\circ C + 273.15 = 292.15 \text{ K}\)
03

Calculate the ideal gas constant

The ideal gas constant (R) is given in SI units as \(8.314 \frac{\text{J}}{\text{mol~K}}\).
04

Find the partial pressure of each gas using the Ideal Gas Law

The Ideal Gas Law states: \(PV = nRT\), where P is pressure, V is volume, n is the amount of gas in moles, R is the ideal gas constant, and T is temperature in Kelvin We will rearrange the Ideal Gas Law to find the partial pressure of each gas: \(P = \frac{nRT}{V}\) Partial pressure of O2: \(P_{O_2} = \frac{n_{O_2}RT}{V} = \frac{1.6 \text{ mol} \cdot 8.314 \frac{\text{J}}{\text{mol~K}} \cdot 292.15 \text{ K}}{10.0 \text{ L}} = 391.54 \text{ Pa}\) Partial pressure of He: \(P_{He} = \frac{n_{He}RT}{V} = \frac{8.15 \text{ mol} \cdot 8.314 \frac{\text{J}}{\text{mol~K}} \cdot 292.15 \text{ K}}{10.0 \text{ L}} = 1987.18 \text{ Pa}\)
05

Determine the total pressure using Dalton's Law of partial pressures

According to Dalton's Law of partial pressures, the total pressure of a mixture of ideal gases is the sum of the partial pressures of its individual gases: \(P_{total} = P_{O_2} + P_{He}\) \(P_{total} = 391.54 \text{ Pa} + 1987.18 \text{ Pa} = 2378.72 \text{ Pa}\) Therefore, the partial pressure of oxygen is \(391.54 \text{ Pa}\), the partial pressure of helium is \(1987.18 \text{ Pa}\), and the total pressure is \(2378.72 \text{ Pa}\).

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Most popular questions from this chapter

An aerosol spray can with a volume of \(125 \mathrm{~mL}\) contains $1.30 \mathrm{~g}\( of propane gas \)\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)$ as a propellant. (a) If the can is at \(25^{\circ} \mathrm{C}\), what is the pressure in the can? (b) What volume would the butane occupy at \(S T P ?\) (c) The can's label says that exposure to temperatures above $50^{\circ} \mathrm{C}$ may cause the can to burst. What is the pressure in the can at this temperature?

(a) Are you more likely to see the density of a gas reported in $\mathrm{g} / \mathrm{mL}, \mathrm{g} / \mathrm{L},\( or \)\mathrm{kg} / \mathrm{cm}^{3} ?(\mathbf{b})$ Which units are appropriate for expressing atmospheric pressures, $\mathrm{N}, \mathrm{Pa}, \mathrm{atm}, \mathrm{kg} / \mathrm{m}^{2} ?$ (c) Which is most likely to be a gas at room temperature and ordinary atmospheric pressure, $\mathrm{F}_{2}, \mathrm{Br}_{2}, \mathrm{~K}_{2} \mathrm{O} .$

A rigid vessel containing a \(3: 1 \mathrm{~mol}\) ratio of carbon dioxide and water vapor is held at \(200^{\circ} \mathrm{C}\) where it has a total pressure of \(202.7 \mathrm{kPa}\). If the vessel is cooled to \(10^{\circ} \mathrm{C}\) so that all of the water vapor condenses, what is the pressure of carbon dioxide? Neglect the volume of the liquid water that forms on cooling.

The physical fitness of athletes is measured by \({ }^{4} V_{\mathrm{O}_{2}}\) max," which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a \(V_{\mathrm{O}_{2}}\) max of \(45 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass \(/ \mathrm{min}\), but a world-class male athlete can have a \(V_{\mathrm{O}_{2}}\) max reading of $88.0 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}$ body mass/min. (a) Calculate the volume of oxygen, in mL, consumed in \(1 \mathrm{hr}\) by an average man who weighs \(85 \mathrm{~kg}\) and has a \(V_{\mathrm{O}_{2}}\) max reading of \(47.5 \mathrm{~mL}\) $\mathrm{O}_{2} / \mathrm{kg}\( body mass \)/ \mathrm{min} .(\mathbf{b})\( If this man lost \)10 \mathrm{~kg},\( exercised, and increased his \)V_{\mathrm{O}_{2}}\( max to \)65.0 \mathrm{~mL} \mathrm{O}_{2} / \mathrm{kg}\( body mass \)/ \mathrm{min}$, how many mL of oxygen would he consume in \(1 \mathrm{hr} ?\)

Rank the following gases from least dense to most dense at $101.33 \mathrm{kPa}\( and \)298 \mathrm{~K}: \mathrm{O}_{2}, \mathrm{Ar}, \mathrm{NH}_{3}, \mathrm{HCl}$.
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