- Does the sealed flask initially contain 40 ml$40\ \mathrm{ml}$ of air or not?
For instance the apparatus could contain two chambers. One 60 ml$60\ \mathrm{ml}$ to be completely filled with liquid acetonitrile and a 40 ml$40\ \mathrm{ml}$ chamber which is evacuated. Then when the experiment starts the chambers are first connected and then heated.
Alternatively the apparatus could contain one 100 ml$100\ \mathrm{ml}$ chamber into which 60 ml of liquid acetonitrile is poured, the other 40 ml$40\ \mathrm{ml}$ being part air and part acetonitrile vapors.
- This is more important if the flask initially contains 40 ml$40\ \mathrm{ml}$ of air.
If yes, then just state the assumption.
If no, then you'll need the partial pressure of acetonitrile at the starting temperature, and you'll have to assume that the 40 ml$40\ \mathrm{ml}$ gaseous phase is in equilibrium with the liquid phase.
- What is the density of liquid(?) acetonitrile at the starting temperature?
Since the problem states "60 ml"$60\ \mathrm{ml}$ acetonitrile" you'd be led to assume that the acetonitrile was initially a liquid.
I'd assume whatever temperature I could find a density for liquid acetonitrile.
- Since 140 °C$140\ \mathrm{^\circ C}$ is above the atmospheric boiling point, which does the acetonitrile do?
If yes, then PV = nRT$pV=nRT$ will work.
If no, then some more "advanced" model needs to be chosen and the additional constants that it needs to better fit the gas phase behavior...
Given the above assumptions, the problem should be easy to solve using PV = nRT$pV=nRT$.
The sealed flask initially contain 40 ml$40\ \mathrm{ml}$ of air.
The starting temperature is $\pu{68 ^\circ F = 20 ^\circ C = 293 ^\circ K}$$\pu{68 ^\circ F = 20 ^\circ C = 293 K}$
The partial pressure of acetonitrile is $\pu{73 mm Hg at 68 ^\circ F}$$\pu{73 mmHg at 68 ^\circ F}$
Density of liquid acetonitrile is $\pu{0.787 g/ml at 68 ^\circ F}$
Since $\pu{140 ^\circ C = 413 ^\circ K}$$\pu{140 ^\circ C = 413 K}$ is above the atmospheric boiling point, the acetonitrile totally vaporizes.
The gaseous acetonitrile acts as if it is a perfect gas.
2 significant figures are needed for the answer.
$$m_{\mathrm{acetonitrile}} = \pu{0.787 g/ml}\times \pu{60 ml} = \pu{47.22 grams}\tag{1}$$$$m_{\text{acetonitrile}} = \pu{0.787 g/ml}\times\pu{60 ml}=\pu{47.22 g}\tag{1}$$
$$n_{\mathrm{acetonitrile}}\dfrac{\pu{47.22 grams}}{\pu{41.053 grams/mole}} = \pu{1.150 moles} \tag{2}$$$$n_{\mathrm{acetonitrile}}\dfrac{\pu{47.22 g}}{\pu{41.053 g/mol}} = \pu{1.150 mol}\tag{2}$$
One mole of a gas at STP occupies $\pu{22.711 L/mol at 0 °C and 100 kPa}$. Since the assumption is that only two significant figures are needed for the answer, the approximately 0.040 L$\pu{22.711 l/mol at $0\ \mathrm{^\circ C}$ and $100\ \mathrm{kPa}$. Since the assumption is that only two significant figures are needed for the answer, the approximately $0.040\ \mathrm l$ of the gas phase (air + acetonitrile) can just be ignored. A more careful check confirms the notion.
$$n_{\mathrm{air}} = \dfrac{PV}{RT} = \dfrac{\pu{100 kPa}\times \pu{0.040 L}}{\pu{8.314 L\cdot kPa} \cdot \pu{^\circ K^{-1}} \times \pu{293 ^\circ K} } = \pu{0.0016 moles}\tag{3}$$$$n_{\text{air}}=\frac{pV}{RT}=\frac{\pu{100 kPa}\times\pu{0.040 l}}{\pu{8.314 l\cdot kPa}\cdot\pu{K^{-1}}\times\pu{293 K}}=\pu{0.0016 mol}\tag{3}$$
$$ n_{\mathrm{liquid\ acetonitrile}} \gg n_{\mathrm{air}} > n_{\mathrm{gaseous\ acetonitrile}}$$$$ n_{\text{liquid acetonitrile}}\gg n_{\text{air}}\gt n_{\text{gaseous acetonitrile}}$$
Now assuming that all of the acetonitrile vaporizes, calculate the pressure assuming that the acetonitrile behaves as an ideal gas equation:
$$P = \dfrac{nRT}{V} = \dfrac{\pu{1.150 moles}\times \pu{8.314 L\cdot kPa} \cdot \pu{^\circ K^{-1}} \cdot \pu{mol^{-1}} \times \pu{413 ^\circ K} }{\pu{0.100 L}}$$
$$ = 39,487 \ \pu{kPa} \ce{->[round]} \pu{3.9\times10^4 kPa} = \pu{3.0\times 10^5 torr}\tag{4}$$$$p=\frac{nRT}V=\frac{\pu{1.150 mol}\times\pu{8.314 l\cdot kPa}\cdot \pu{K^{-1}}\cdot\pu{mol^{-1}}\times\pu{413 K}}{\pu{0.100 l}}\\ =\pu{39 487 kPa}\approx\pu{3.9\times10^4 kPa}=\pu{3.0\times 10^5 Torr}\tag{4}$$
On the Wikipedia data page for acetonitrile gives a formula for the vapor pressure (Temperaturetemperature range from $\pu{229.32 ^\circ K}$$\pu{229.32 K}$ to $\pu{545.50 ^\circ K}$$\pu{545.50 K}$) where P$p$ is the pressure in mm Hg$\mathrm{mmHg}$ and T$T$ is the temperature in $^\circ \pu{K}$$\mathrm{K}$:
$$\ln{\pu{P}} = \ln{\dfrac{760}{101.325}} -3.881710\cdot \ln{\pu{T}} - \dfrac{4999.618}{\pu{T}} + 41.05901 + 3.155956\cdot10^{-6}\cdot\pu{T}^2$$$$\ln p=\ln{\dfrac{760}{101.325}}-3.881710\cdot \ln T-\dfrac{4999.618}T+41.05901+3.155956\cdot10^{-6}\cdot T^2$$
$$\therefore \pu{P} = \pu{3,379 torr} \ce{->[round]} \pu{3.4\cdot 10^{3} torr}$$$$\therefore p=\pu{3 379 Torr}\approx\pu{3.4\cdot10^{3} Torr}$$
Since the vapor pressure of acetonitrile in equilibrium with the liquid at $\pu{413 ^\circ K}$$\pu{413 K}$ is less than the vapor pressure that would be created by all of the acetonitrileacetonitrile evaporating, all of the acetonitrile will not evaporate. The relative volumes of the two phases won't matter for the pressure.
