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 \(130^{\circ}\) F may cause the can to burst. What is the pressure in the can at this temperature?

First, we need to convert the gas mass (2.30 g) to moles using propane's molar mass: Molar mass of propane (C3H8) = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol Number of moles (n) = (2.30 g) / (44.11 g/mol) = 0.0521 mol Next, we need to convert the given temperature (23°C) to Kelvin: T = 23°C + 273.15 = 296.15 K Now, let's find the pressure using the Ideal Gas Law. We need to convert the volume to Liters first: V = 250 mL / 1000 = 0.250 L The ideal gas constant (R) in L·atm/mol·K = 0.0821 L·atm/mol·K Now, we plug the values into the Ideal Gas Law to find the pressure (P): PV = nRT P = nRT / V P = (0.0521 mol)(0.0821 L·atm/mol·K)(296.15 K) / (0.250 L) P = 5.09 atm The pressure in the can is 5.09 atm at 23°C.

Standard Temperature and Pressure (STP) conditions are 0°C (273.15 K) and 1 atm. Using the Ideal Gas Law, we can calculate the volume at STP: \(V_{STP} = \frac{nRT_{STP}}{P_{STP}}\) \(V_{STP} = \frac{(0.0521 \text{ mol})(0.0821 \text{ L·atm/mol·K})(273.15 \text{ K})}{1 \text{ atm}}\) \(V_{STP} = 1.18 \text{ L}\) At STP, the propane gas occupies a volume of 1.18 L.

First, convert the given temperature (130°F) to Kelvin: \(T_{new} = \frac{5}{9}(130 - 32) + 273.15 = 324.15 K\) Now, apply the Ideal Gas Law to find the pressure in the can at this new temperature: \(P_{new} = \frac{nRT_{new}}{V}\) \(P_{new} = \frac{(0.0521 \text{ mol})(0.0821 \text{ L·atm/mol·K})(324.15 \text{ K})}{0.250 \text{ L}}\) \(P_{new} = 7.07 \text{ atm}\) At a temperature of 130°F, the pressure in the can is 7.07 atm.