EDIT I fouled up my original answer. So as a matter of pride I have fixed it. The OP was right on his idea #1. The answer to the question hinges on the calculated vapor pressure assuming all of the acetonitrile evaporates. I have no idea what coefficients the OP might have had for the Antoine equation.
Let's go down the rabbit hole! Given no other information you have to make several assumptions.
- Does the sealed flask initially contain $40\ \mathrm{ml}$ of air or not?
For instance the apparatus could contain two chambers. One $60\ \mathrm{ml}$ to be completely filled with liquid acetonitrile and a $40\ \mathrm{ml}$ chamber which is evacuated. Then when the experiment starts the chambers are first connected and then heated.
Alternatively the apparatus could contain one $100\ \mathrm{ml}$ chamber into which 60 ml of liquid acetonitrile is poured, the other $40\ \mathrm{ml}$ being part air and part acetonitrile vapors.
- What is the starting temperature?
- This is more important if the flask initially contains $40\ \mathrm{ml}$ of air.
- Can the partial pressure of acetonitrile at the starting temperature be ignored?
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 $40\ \mathrm{ml}$ gaseous phase is in equilibrium with the liquid phase.
- What is the density of liquid(?) acetonitrile at the starting temperature?
Since the problem states "$60\ \mathrm{ml}$ 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 $140\ \mathrm{^\circ C}$ is above the atmospheric boiling point, which does the acetonitrile do?
- Does gaseous acetonitrile act as if it is a perfect gas?
If yes, then $pV=nRT$ 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.
- Number of significant figures for the answer?
- It would seem to be 2, so this feeds back into assumptions 2, 3, 4, and 6.
Given the above assumptions, the problem should be easy to solve using $pV=nRT$.
Let's assume:
The sealed flask initially contain $40\ \mathrm{ml}$ of air.
The starting temperature is $\pu{68 ^\circ F = 20 ^\circ C = 293 K}$
The partial pressure of acetonitrile is $\pu{73 mmHg at 68 ^\circ F}$
Density of liquid acetonitrile is $\pu{0.787 g/ml at 68 ^\circ F}$
Since $\pu{140 ^\circ C = 413 K}$ 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.
OK, let's check some our our assumptions. For the acetonitrile:
$$m_{\text{acetonitrile}} = \pu{0.787 g/ml}\times\pu{60 ml}=\pu{47.22 g}\tag{1}$$
$$n_{\mathrm{acetonitrile}}\dfrac{\pu{47.22 g}}{\pu{41.053 g/mol}} = \pu{1.150 mol}\tag{2}$$
One mole of a gas at STP occupies $\pu{22.711 l/mol at $0\ \mathrm{^\circ C}$ 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.
$$n_{\text{air}}=\frac{pV}{RT}=\frac{\pu{100 kPa}\times\pu{0.040 l}}{\pu{8.314 l\cdot kPa}\cdot\pu{K^{-1}}\times\pu{293 K}}=\pu{0.0016 mol}\tag{3}$$
$$ n_{\text{liquid acetonitrile}}\gg n_{\text{air}}\gt n_{\text{gaseous acetonitrile}}$$
Now assuming that all of the acetonitrile vaporizes, calculate the pressure assuming that the acetonitrile behaves as an ideal gas equation:
$$p=\frac{nRT}V=\frac{\pu{1.150 mol}\times\pu{8.314 l\cdot kPa}\cdot \pu{K^{-1}}\cdot\pu{mol^{-1}}\times\pu{413 K}}{\pu{0.100 l}}\\
=\pu{39 487 kPa}\approx\pu{3.9\times10^4 kPa}=\pu{3.0\times 10^5 Torr}\tag{4}$$
On the Wikipedia data page for acetonitrile gives a formula for the vapor pressure (temperature range from $\pu{229.32 K}$
to $\pu{545.50 K}$) where $p$ is the pressure in $\mathrm{mmHg}$ and $T$ is the temperature in $\mathrm{K}$:
$$\ln p=\ln{\dfrac{760}{101.325}}-3.881710\cdot \ln T-\dfrac{4999.618}T+41.05901+3.155956\cdot10^{-6}\cdot T^2$$
$$\therefore p=\pu{3 379 Torr}\approx\pu{3.4\cdot10^{3} Torr}$$
Since the vapor pressure of acetonitrile in equilibrium with the liquid at $\pu{413 K}$ is less than the vapor pressure that would be created by all of the acetonitrile evaporating, all of the acetonitrile will not evaporate. The relative volumes of the two phases won't matter for the pressure.
