/*! 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 40 An aerosol spray can with a volu... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 40

An aerosol spray can with a volume of \(250 \mathrm{~mL}\) contains \(2.30 \mathrm{~g}\) of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) as a propellant. (a) If the can is at \(23{ }^{\circ} \mathrm{C}\), what is the pressure in the can? (b) What volume would the propane occupy at STP? (c) The can's label says that exposure to temperatures above \(130^{\circ} \mathrm{F}\) may cause the can to burst. What is the pressure in the can at this temperature?

Short Answer

Expert verified
The pressure in the aerosol spray can at \(23^{\circ}C\) is \(5.11\,atm\). The volume of propane at STP is \(1.16\,L\). The pressure in the can at \(130^{\circ}F\) is \(6.86\,atm\).

Step by step solution

01

(a) Finding the pressure at 23掳C:

1. Convert the temperature to Kelvin: \(T(K) = T(掳C) + 273.15 = 23 + 273.15 = 296.15 \mathrm{K}\). 2. Convert the mass of propane into moles: Molar mass of propane = \(3 * \text{mass of carbon} + 8 * \text{mass of hydrogen} = 3 * 12.01 + 8 * 1.01 = 44.11 \mathrm{g/mol}\). Number of moles (n) = \(\frac{\text{mass}}{\text{molar mass}} = \frac{2.30 \mathrm{g}}{44.11 \mathrm{g/mol}} = 0.0521 \mathrm{mol}\). 3. Find the volume in liters: Since the volume given is in milliliters, we'll convert it into liters: \(250 \mathrm{mL} * \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.250 \mathrm{L}\). 4. Use the Ideal Gas Law to find the pressure: \(PV = nRT\) We need to find P, so we can rewrite this as: \(P = \frac{nRT}{V}\) Using the value for the universal gas constant, \(R = 0.0821 \mathrm{L} * \mathrm{atm} / (\mathrm{mol} * \mathrm{K})\), we obtain: \(P = \frac{(0.0521 \mathrm{mol})(0.0821 \mathrm{L} * \mathrm{atm} / (\mathrm{mol} * \mathrm{K}))(296.15 \mathrm{K})}{0.250L} = 5.11 \mathrm{atm}\) So, the pressure in the can at 23掳C is 5.11 atm.
02

(b) Finding the volume of propane at STP:

1. To find the volume at STP, we will use the Ideal Gas Law again. We are given that the temperature at STP is 0掳C (273.15K) and the pressure is 1 atm. 2. Use the Ideal Gas Law with the new values(substitute P=1, T=273.15): \(V = \frac{nRT}{P} = \frac{(0.0521 \mathrm{mol})(0.0821 \mathrm{L} * \mathrm{atm} / (\mathrm{mol} * \mathrm{K}))(273.15\mathrm{K})}{1 \mathrm{atm}} = 1.16 \mathrm{L}\) So, the volume of propane at STP is 1.16 L.
03

(c) Finding the pressure at 130掳F:

1. Convert the temperature from Fahrenheit to Kelvin: \(T(掳C) = \frac{5}{9}(T(掳F) - 32) = \frac{5}{9} (130-32) = 54.44 \mathrm{掳C}\) \(T(K) = T(掳C) + 273.15 = 54.44 + 273.15 = 327.59 \mathrm{K}\) 2. Use the Ideal Gas Law to find the pressure: \(P =\frac{nRT}{V}\) Using the values calculated previously, and the new temperature of 327.59 K, we find: \(P =\frac{(0.0521 \mathrm{mol})(0.0821 \mathrm{L} * \mathrm{atm} / (\mathrm{mol} * \mathrm{K}))(327.59 \mathrm{K})}{0.250 \mathrm{L}} = 6.86 \mathrm{atm}\) The pressure in the can at 130掳F is 6.86 atm.

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Most popular questions from this chapter

A 6.53-g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces \(1.72 \mathrm{~L}\) of carbon dioxide gas at \(28^{\circ} \mathrm{C}\) and 743 torr pressure. (a) Write balanced chemical equations for the reactions that occur between hydrochloric acid and each component of the mixture. (b) Calculate the total number of moles of carbon dioxide that forms from these reactions. (c) Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

The Goodyear blimps, which frequently fly over sporting events, hold approximately \(175,000 \mathrm{ft}^{3}\) of helium. If the gas is at \(23^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\), what mass of helium is in a blimp?

Perform the following conversions: (a) \(0.912\) atm to torr, (b) \(0.685\) bar to kilopascals, (c) \(655 \mathrm{~mm} \mathrm{Hg}\) to atmospheres, (d) \(1.323 \times 10^{5} \mathrm{~Pa}\) to atmospheres, (e) \(2.50 \mathrm{~atm}\) to psi.

A quantity of \(\mathrm{N}_{2}\) gas originally held at \(5.25 \mathrm{~atm}\) pressure in a \(1.00\) - \(\mathrm{L}\) container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5\) - \(\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(5.25\) atm and \(26^{\circ} \mathrm{C}\) in a \(5.00\)-L container is transferred to this same container. What is the total pressure in the new container?

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. (a) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at \(0.984 \mathrm{~atm}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4}\) ?
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