Question

a spray can is used until it is empty except for the propellant gas, which has a pressure of 1340 torr at 23°c. if the can is thrown into a fire (t = 475°c), what will be the pressure in the hot can? which gas law equation should be used to solve this problem? ( pv = nrt ) ( p_1v_1 = p_2v_2 ) ( \frac{v_1}{t_1} = \frac{v_2}{t_2} ) ( \frac{p_1v_1}{t_1} = \frac{p_2v_2}{t_2} ) ( \frac{v_1}{n_1} = \frac{v_2}{n_2} ) ( \frac{p_1}{t_1} = \frac{p_2}{t_2} ) solve for the unknown variable. torr

Explanation:

Step1: Identify the gas law

Since the volume \(V\) and amount of gas \(n\) are constant (the can's volume doesn't change and the amount of propellant gas is fixed as the can is "empty except for propellant"), we use Gay - Lussac's law \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\).

Step2: Convert temperatures to Kelvin

\(T_1=(23 + 273)\text{ K}=296\text{ K}\), \(T_2=(475+ 273)\text{ K}=748\text{ K}\), \(P_1 = 1340\text{ torr}\).
From \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\), we can solve for \(P_2\):
\(P_2=\frac{P_1T_2}{T_1}\)
Substitute the values: \(P_2=\frac{1340\times748}{296}\)
\(P_2=\frac{1340\times748}{296}=1340\times\frac{748}{296}\)
\(\frac{748}{296} = 2.53\)
\(P_2=1340\times2.53 = 3390.2\text{ torr}\)

Answer:

\(\frac{P_1}{T_1}=\frac{P_2}{T_2}\), \(3390\) torr (rounded to three significant figures)