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?

The pressure in the can at 25°C is found using the Ideal Gas Law equation rearranged for P and plugging in the calculated values of n, V, and T: \[P = \frac{(1.3\mathrm{~g}/44.097\mathrm{~g/mol})\times(8.314\mathrm{~J/[mol\cdot K]})\times(298.15\mathrm{~K})}{125\mathrm{~mL}}\] The volume of propane at STP is calculated using the Ideal Gas Law equation rearranged for V and plugging in the values for n, R, and T: \[V_{STP} = \frac{(1.3\mathrm{~g}/44.097\mathrm{~g/mol})\times(8.314\mathrm{~J/[mol\cdot K]}/101.325\mathrm{~J/[L\cdot atm]})\times(273.15\mathrm{~K})}{1\mathrm{~atm}}\] Lastly, the pressure in the can at 50°C is calculated using the Ideal Gas Law equation rearranged for P and plugging in the calculated values of n, V, and T': \[P' = \frac{(1.3\mathrm{~g}/44.097\mathrm{~g/mol})\times(8.314\mathrm{~J/[mol\cdot K]})\times(323.15\mathrm{~K})}{125\mathrm{~mL}}\]