Answer:
About 3.81 grams.
Explanation:
We can use the ideal gas law. Recall that:
[tex]\displaystyle PV = nRT[/tex]
Note that the universal gas constant R has the value 0.08206 L-atm/mol-K.
Hence, convert the measured pressure to atms (1 atm = 14.7 psi):
[tex]\displaystyle 27.0\text{ psi} \cdot \frac{1\text{ atm}}{14.7\text{ psi}} = 1.84\text{ atm}[/tex]
Rearrange the equation to solve for n, the number of moles of vaporized triethylamine and evaluate. The temperature is (25.0 + 273.15 ) K = 298.2 K:
[tex]\displaystyle \begin{aligned} n& = \frac{PV}{RT} \\ \\ & = \frac{(1.84\text{ atm})(0.500\text{ L})}{\left(\dfrac{0.08206\text{ L - atm}}{\text{mol-K}}\right)(273.2\text{ K})} \\ \\ &= 0.0376\text{ mol N(CH$_2$CH$_3$)$_3$}\end{aligned}[/tex]
Convert from moles to grams:
[tex]\displaystyle 0.0376\text{ mol N(CH$_2$CH$_3$)$_3$} \cdot \frac{101.22\text{ g N(CH$_2$CH$_3$)$_3$}}{1\text{ mol N(CH$_2$CH$_3$)$_3$}} = 3.81\text{ g N(CH$_2$CH$_3$)$_3$}}[/tex]
In conclusion, there is about 3.81 grams of vaporized triethylamine in the lecture bottle.
