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Example 9.5: Predicting Change in Pressure with Temperature

A can of hair spray is used until it is empty except for the propellant, isobutane gas.
(a) On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why?
(b) The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can?

Solution

\(T_{\mathrm{1}}\) \(= 24.\ \mathrm{°aC}\)


\(V_{\mathrm{gas}}\) \(= 350.\ \mathrm{mL}\)


\(P_{\mathrm{1}}\) \(= 360.\ \mathrm{kPa}\)


\(P_{\mathrm{2}}\) = ?


\(T_{\mathrm{2}}\) \(= 50.\ \mathrm{°aC}\)


(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)
(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’s/Gay-Lussac’s law. Taking \(\ce{P1}\) and \(T_{\mathrm{1}}\) as the initial values, \(T_{\mathrm{2}}\) as the temperature where the pressure is unknown and \(\ce{P2}\) as the unknown pressure, we have:


which means that

\(P_{\mathrm{2}}\) \(= P_{\mathrm{1}} \cdot \dfrac{T_{\mathrm{2}}}{T_{\mathrm{1}}}\)

\(\ \ \ =360.\ \mathrm{kPa} \cdot \dfrac{323.\ \mathrm{K}}{297.\ \mathrm{K}}\)

\(\ \ \ =360.\ \mathrm{kPa} \cdot 1.0875\)

\(\ \ \ =391.\ \mathrm{kPa}\)