Problem 40 - Chemistry: The Central Science

Question

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?

Step-by-Step Solution

Answer

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 1: (a) Finding the pressure at 23°C:

1. Convert the temperature to Kelvin: \(T(K) = T(°C) + 273.15 = 23 + 273.15 = 296.15 \mathrm{K}\). 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°C is 5.11 atm.

Step 2: (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°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.

Step 3: (c) Finding the pressure at 130°F:

1. Convert the temperature from Fahrenheit to Kelvin: \(T(°C) = \frac{5}{9}(T(°F) - 32) = \frac{5}{9} (130-32) = 54.44 \mathrm{°C}\) \(T(K) = T(°C) + 273.15 = 54.44 + 273.15 = 327.59 \mathrm{K}\) 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°F is 6.86 atm.