# Filling up a tank with a liquid that has an insoluble, inert gas insid

member 392791

## The Attempt at a Solution

The attempt at the solution has been attached, I'm up to part C on this though, so I haven't attempted d or f. I'll get to those as soon as part c is clear as well as A and B.

I don't understand how doing a gas phase mass balance will help, and I'm not even sure how to perform a mass balance on the gas, but I did attempt it.

I'm not sure if I'm doing in the right direction with this problem

#### Attachments

• Problem 2.8.pdf
40.7 KB · Views: 457
• Problem 2.8 attempt.pdf
229.1 KB · Views: 181
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SteamKing
Staff Emeritus
Homework Helper
Seems like the bottom of your page of calculations got cut off.

Chestermiller
Mentor
Woopydalan,

Your assessment is right. There is no reason to do a mass balance on the gas. You are asked to find the pressure in the tank as a function of the liquid volume. Let V be the liquid volume. Then the volume of the gas is V0-V. You can then use the ideal gas law to express the pressure in the gas as a function of the liquid volume. You can also use this equation to find the final liquid volume.

For the last part, since you don't know the proportionality constant between the liquid flow rate and the pressure difference, just call it k:

$$\frac{dV}{dt}=k(p_{in}-p)$$

Substitute your equation for p in terms of V into this relationship.

member 392791
How does this look?

Also, here is my attempt at part d and e. Part E says to perform a mass balance on the liquid, is that even possible?

Suppose this was a test question, since I never did the mass balances I would look a lot of credit I assume, but maybe my professor made a typo or doesn't know that you can't mass balance it??

#### Attachments

• 2.8 attempt 2.pdf
227 KB · Views: 149
• 2.8 attempt part d and e.pdf
232.6 KB · Views: 166
Last edited by a moderator:
Chestermiller
Mentor
Since this is an isothermal problem, let's lose the nRT. So, if V = V(t) is the volume of liquid in the tank at time t, the volume of gas in the tank at time t is V0-V. From the ideal gas law:
$$P(V_0-V)=P_0V_0=P_{in}(V_0-V_f)$$
where Vf is the final volume of liquid in the tank. From the above equation, you get:
$$V_f=V_0\left(1-\frac{P_0}{P_{in}}\right)$$
and
$$P=\frac{P_0V_0}{(V_0-V)}$$
Since
$$\frac{dV}{dt}=k(P_{in}-P)$$
we can substitute for P and obtain:
$$\frac{dV}{dt}=k\left(P_{in}-\frac{P_0V_0}{(V_0-V)}\right)$$

I believe this is the relationship they were looking for for part e. Some of these results are equivalent to what you already obtained in your attachments. I just wanted to demonstrate an easier way of getting the results.

Chet

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