39 grams of benzene (z = 1.4) is filled in a container of volume 10 liter at 27°C, what is the pressure of container?

Question

Quizplus · Accepted Answer

Using the ideal gas equation, $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant (0.0821 L atm K$^{-1}$ mol$^{-1}$), and $T$ is the temperature in Kelvin. First, convert the temperature to Kelvin: $T = 27°C + 273 = 300K$. Calculate the number of moles of benzene: the molar mass of benzene (C$_6$H$_6$) is $78 g/mol$, so $n = 39g / 78g/mol = 0.5 mol$. Substitute the values into the ideal gas equation and solve for $P$: $P = (nRT)/V = (0.5 mol \times 0.0821 L atm K$^{-1}$ mol$^{-1}$ \times 300K) / 10L = 1.23 atm$. However, benzene is not an ideal gas, and the question provides a compressibility factor $z = 1.4$, indicating deviation from ideal behavior. To correct for non-ideal behavior, multiply the pressure by $z$: $P_{real} = P_{ideal} \times z = 1.23 atm \times 1.4 = 1.722 atm$, which rounds to $1.7 atm$, making option A correct.

39 Grams of Benzene (Z = 1