# Antoine law and ideal gas law

I have the following question:

A $$100\ \mathrm{ml}$$ sealed flask with $$60\ \mathrm{ml}$$ acetonitrile is heated to $$140\ \mathrm{^\circ C}$$. The boiling point of acetonitrile is $$82\ \mathrm{^\circ C}$$. What will be the pressure in the flask?

1) I understand I can use the Antoine equation, $$\log{p} = A-B/(C+T)$$, to calculate the vapor pressure of acetonitrile, but that equation only applies if there is some liquid acetonitrile left in the flask. How do I know if this is the case, or instead all acetonitrile has boiled and is in the vapor phase?

2) The total pressure is the sum of the partial pressures of acetonitrile and air. I can calculate the air pressure inside the flask with the ideal gas law, PV=nRT. However, I'm not sure what is the volume? Is it the total volume minus the acetonitrile volume ($$100\ \mathrm{ml}-60\ \mathrm{ml}=40\ \mathrm{ml}$$), or is it $$40\ \mathrm{ml}+x$$, where $$x$$ is the additional volume due to the conversion of liquid acetonitrile into vapor?

Any ideas are welcome.

• To know the partial pressure of the air in the flask following heating you need to know how many moles of air you have, which you can calculate from the partial pressure of air at the start.But to determine this, you need the initial temp. of the flask, and the conditions under which it was sealed (e.g., was it sealed such that tthere was 60 mL acetonitrile after saturating the air with it, or was it sealed before the acetonitrile vaporized, in which case the partial pressure of the air would be 1 atm)? Or did it have any air to start? You need this info. to begin answering the question. Commented Apr 2, 2020 at 7:03
• Ad 1) Remember the pressurized cooking. The water is filling the comparable fraction of the volume, and temperature is raised above the standard boiling point by comparable amount. No cook is afraid of total vaporization. Commented Apr 2, 2020 at 8:42
– user91590
Commented Apr 2, 2020 at 13:39

Assume that the initial temperature is room temperature and neglect the initial amount of acetonitrile in the gas phase. Calculate the equilibrium vapor pressure of acetonitrile at 140 C and compare it with the partial pressure you would calculate from the ideal gas law if all the acetonitrile had evaporated (so that its volume is 100 cc). If the latter is greater than the former, then there will still be some acetonitrile remaining as liquid. The partial pressure of the acetonitrile in the gas phase will then be equal to the equilibrium vapor pressure. Assuming that x gm of acetonitrile have evaporated, you can then calculate, in terms of x, the new mass and volume of the liquid and the new volume of the head space. You can then determine x from the requirement that the total amount of acetonitrile has not changed.

• Thank you, that was very helpful. As the partial pressure calculated with the ideal gas law is greater, I can use the Antoine equation and solve the problem.
– user91590
Commented Apr 2, 2020 at 13:36

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.

1. 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.

1. What is the starting temperature?
• This is more important if the flask initially contains $$40\ \mathrm{ml}$$ of air.
1. 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.

1. 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.

1. Since $$140\ \mathrm{^\circ C}$$ is above the atmospheric boiling point, which does the acetonitrile do?
• Totally vaporize?

• Form gas and a liquid phases?

• Turn into a supercritical fluid?

1. 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.

1. 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:

1. The sealed flask initially contain $$40\ \mathrm{ml}$$ of air.

2. The starting temperature is $$\pu{68 ^\circ F = 20 ^\circ C = 293 K}$$

3. The partial pressure of acetonitrile is $$\pu{73 mmHg at 68 ^\circ F}$$

4. Density of liquid acetonitrile is $$\pu{0.787 g/ml at 68 ^\circ F}$$

5. Since $$\pu{140 ^\circ C = 413 K}$$ is above the atmospheric boiling point, the acetonitrile totally vaporizes.

6. The gaseous acetonitrile acts as if it is a perfect gas.

7. 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.

• 1. It is one 100 ml chamber. 2. I agree, so I have to assume the starting temp. 3. I don't know. 5+6. That are the problem that concerns me most, because I don't know it.
– user91590
Commented Apr 2, 2020 at 13:34