/*! 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 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
(a) The pressure in the can at \(23^{\circ} \mathrm{C}\) is \(5.18 \mathrm{~atm}\). (b) The volume of propane at STP is \(1.12 \mathrm{~L}\). (c) The pressure in the can at \(130^{\circ} \mathrm{F}\) is \(6.89 \mathrm{~atm}\).

Step by step solution

01

Calculate the number of moles of propane (C3H8)

Since the mass of propane in the can is given, we need to calculate the number of moles of propane. To do this, we find the molar mass of propane and then divide the mass by the molar mass. Molar mass of C3H8 = \((3 * 12.01) + (8 * 1.01) = 44.11 \mathrm{~g/mol}\) Number of moles, \(n = \frac{2.30 \mathrm{~g}}{44.11 \mathrm{~g/mol}} = 0.0521 \mathrm{~mol}\)
02

(a) Calculate the pressure in the can

Now, we can use the Ideal Gas Law to calculate the pressure in the can. The given temperature must be converted to Kelvin (K): \(T = 23^{\circ} \mathrm{C} + 273.15 = 296.15 \mathrm{~K}\) Volume of the can is given as \(250 \mathrm{~mL}\), which needs to be converted to liters: \(V = 250 \mathrm{~mL} = 0.250 \mathrm{~L}\) Using the Ideal Gas Law, \(PV = nRT\), we can solve for pressure \(P\): \(P = \frac{nRT}{V} = \frac{0.0521 \mathrm{~mol} * 0.0821 \mathrm{~L \cdot atm/mol \cdot K} * 296.15 \mathrm{~K}}{0.250 \mathrm{~L}} = 5.18 \mathrm{~atm}\) So, the pressure in the can is \(5.18 \mathrm{~atm}\).
03

(b) Calculate the volume of propane at STP

The volume of the propane at STP can also be calculated using the Ideal Gas Law. Here, \(T = 273.15 \mathrm{~K}\) (0 掳C) and the standard pressure is \(P = 1 \mathrm{~atm}\). Using the Ideal Gas Law, we can solve for the volume, \(V\): \(V = \frac{nRT}{P} = \frac{0.0521 \mathrm{~mol} * 0.0821 \mathrm{~L \cdot atm/mol \cdot K} * 273.15 \mathrm{~K}}{1 \mathrm{~atm}} = 1.12 \mathrm{~L}\) So, the volume of propane at STP is \(1.12 \mathrm{~L}\).
04

(c) Calculate the pressure in the can at 130 掳F

In this case, we need to convert the given temperature, \(130^{\circ} \mathrm{F}\), to Kelvin. To do that, first, convert 掳F to 掳C using the following formula: \(T_{掳C} = (T_{掳F} - 32) * 5 / 9 \). Then, convert 掳C to Kelvin (K) by adding 273.15. \(T = (130 - 32) * \frac{5}{9} \mathrm{~掳C} = 54.44 \mathrm{~掳C}\) \(T = 54.44 \mathrm{~掳C} + 273.15 = 327.59 \mathrm{~K}\) Using the Ideal Gas Law (\(PV = nRT\)), we can solve for the pressure in the can at this temperature: \(P = \frac{nRT}{V} = \frac{0.0521 \mathrm{~mol} * 0.0821 \mathrm{~L \cdot atm/mol \cdot K} * 327.59 \mathrm{~K}}{0.250 \mathrm{~L}} = 6.89 \mathrm{~atm}\) So, the pressure in the can at \(130^{\circ} \mathrm{F}\) is \(6.89 \mathrm{~atm}\).

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Most popular questions from this chapter

To minimize the rate of evaporation of the tungsten filament, \(1.4 \times 10^{-5}\) mol of argon is placed in a \(600-\mathrm{cm}^{3}\) lightbulb. What is the pressure of argon in the lightbulb at \(23^{\circ} \mathrm{C} ?\)

After the large eruption of Mount St. Helens in 1980 , gas samples from the volcano were taken by sampling the downwind gas plume. The unfiltered gas samples were passed over a gold-coated wire coil to absorb mercury (Hg) present in the gas. The mercury was recovered from the coil by heating it, and then analyzed. \(\underline{\text { In one }}\) particular set of experiments scientists found a mercury vapor level of \(1800 \mathrm{ng}\) of \(\mathrm{Hg}\) per cubic meter in the plume, at a gas temperature of \(10^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of \(\mathrm{Hg}\) vapor in the plume, \((\mathrm{b})\) the number of \(\mathrm{Hg}\) atoms per cubic meter in the gas, \((\mathrm{c})\) the total mass of Hg emitted per day by the volcano if the daily plume volume was \(1600 \mathrm{~km}^{3}\).

The density of a gas of unknown molar mass was measured as a function of pressure at \(0{ }^{\circ} \mathrm{C}\), as in the table below. (a) Determine a precise molar mass for the gas. Hint: Graph \(d / P\) versus \(P\). (b) Why is \(d / P\) not a constant as a function of pressure? $$ \begin{array}{llllll} \hline \begin{array}{l} \text { Pressure } \\ \text { (atm) } \end{array} & 1.00 & 0.666 & 0.500 & 0.333 & 0.250 \\ \text { Density } & & & & & \\ \text { (g/L) } & 2.3074 & 1.5263 & 1.1401 & 0.7571 & 0.5660 \\ \hline \end{array} $$

A fixed quantity of gas at \(21{ }^{\circ} \mathrm{C}\) exhibits a pressure of 752 torr and occupies a volume of \(4.38 \mathrm{~L}\). (a) Use Boyle's law to calculate the volume the gas will occupy if the pressure is increased to \(1.88 \mathrm{~atm}\) while the temperature is held constant. (b) Use Charles's law to calculate the volume the gas will occupy if the temperature is increased to \(175^{\circ} \mathrm{C}\) while the pressure is held constant.

Based on their respective van der Waals constants (Table 10.3), is Ar or \(\mathrm{CO}_{2}\) expected to behave more nearly like an ideal gas at high pressures? Explain.
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