# Pressure of a fluid contained in a steel container

1. Apr 26, 2015

### squelch

1. The problem statement, all variables and given/known data

A steel container is completely filled with gasoline, and then sealed. The temperature is then increased 11 degrees C. If the bulk modulus of gasoline is $1.90 \times {10^9}Pa$, find the increase in pressure of the gasoline if:
(a) The expansion of the container is considered.
(b) The expansion of the container is ignored.
(For gasoline, $\beta = 9.60 \times {10^{ - 4}}^ \circ {C^{ - 1}}$)

2. Relevant equations

$$\Delta V = {V_0}\beta \Delta T$$
$$p = \frac{F}{A} = - Y\alpha \Delta T$$

3. The attempt at a solution

The volume of the steel container will expand to a final volume:
$${V_c} = {V_0}(1 + {\beta _{(steel)}}\Delta T)$$
Similarly, the volume of the gas expands:
$${V_c} = {V_0}(1 + {\beta _{(gasoline)}}\Delta T)$$

Because we filled the container up to the brim before sealing it, we can assume the initial volumes are equal and that the initial pressure is one atmosphere (how relevant the latter point is I'm not sure). The initial volume isn't given, but I can relate the two ratios:

$$\frac{{\Delta {V_s}}}{{{\beta _s}}} = \frac{{\Delta {V_g}}}{{{\beta _g}}}$$

Presumably, the liquid gasoline expands more than the steel canister. I know that pressure = force / area and I know that $p = \frac{F}{A} = - Y\alpha \Delta T$, but I'm not entirely sure how to apply that latter equation to what I know.

2. Apr 26, 2015

### Staff: Mentor

The easiest approach is to treat this as a two-step process. For the case where the container expansion is ignored, first calculate the new volume of the gasoline as if it were completely free to expand (thermal expansion). Then calculate the pressure increase that would be required to squeeze the gasoline back into the container (bulk modulus).

Chet

3. Apr 26, 2015

### squelch

I think my *biggest* problem is that I can't seem to find an expression for the pressure increase required to compress the gasoline back into the container. Obviously this isn't an ideal gas, so I can't use the Ideal Gas Law. I'm not really sure at all what the $-Y\alpha \Delta T$ is trying to say. I just found an expression for the bulk modulus:
$$B = \frac{{\Delta P}}{{\frac{{\Delta V}}{V}}} = \frac{{\Delta P \cdot V}}{{\Delta V}}$$
...and I'm trying to proceed from there.

4. Apr 26, 2015

### Staff: Mentor

I don't know what the Y in your equation stands for, but I'm guessing it is related to the ratio of the coefficient of thermal expansion to the bulk modulus. As far as your approach that you are trying to proceed with, this is the correct way to go.

Chet

5. Apr 27, 2015

### squelch

So I found a procedure and tried to implement it this way:

$$\begin{array}{l} \Delta {V_{(container)}} = \Delta {V_{(therm.expan.)}} - \Delta {V_{(stress)}}\\ ⇒ {V_0}{\beta _{(steel)}}\Delta T = {V_0}{\beta _{gasoline}} - \frac{{\Delta P \cdot {V_0}}}{{{B_{gasoline}}}}\\ ⇒ ({\beta _{gasoline}} - {\beta _{steel}})\Delta T{B_{gasoline}} = \Delta P \end{array}$$

Plugging in the known constants, and taking the bulk modulus of steel to be $3.6 \times {10^{ - 5}}^o{C^{ - 1}}$ I get an answer of $1.756 \times {10^7}Pa$ for the first portion. Does this seem sensible?

6. Apr 27, 2015

### Delta²

Thats what i also had in mind. b) is the same procedure, with just taking $\Delta V_{container}=0$

7. Apr 27, 2015

### squelch

Yup, that's the process I'm applying to the latter half. Thanks for the validation.

8. Apr 27, 2015

### squelch

As a last point, I'm actually finding $\Delta P$, so is it accurate to say I need to add 1 ATM to the pressure?

edit: Upon rereading the question it's clear that this is the pressure reading they want, but I suppose it's worth knowing that the absolute pressure is $\Delta P + 1 atm$

Last edited: Apr 27, 2015