# PhysicsAverage Density of Gasoline

#### Cbarker1

##### Active member
Dear Everybody,

A 2.80 kg steel gas can holds 20 L of gasoline when full. What's the average density (in kg/m^3) of full gas can, taking into the volume occupied by steel as well as by gasoline?

Work:

Given
the mass of the steel gas can= 2.80 kg
The Total Volume of Gas can= 20.0 L
The density of the gasoline= 680 kg/m^3

$\rho=\frac{m}{Volume of the gasoline}$

How to setup the problem?

Thanks
Cbarker1

#### MarkFL

Staff member
Obviously, if the can holds 20.0 L of gasoline, then the total volume of the can must be greater than 20.0 L. Let's assume that the can is made of stainless steel with a mass density of 8000 kg/m³. What would the volume of the given mass of steel be?

#### Cbarker1

##### Active member
Obviously, if the can holds 20.0 L of gasoline, then the total volume of the can must be greater than 20.0 L. Let's assume that the can is made of stainless steel with a mass density of 8000 kg/m³. What would the volume of the given mass of steel be?
the volume is 2857.14 m^3.

#### MarkFL

Staff member
the volume is 2857.14 m^3.
I think you have the volume inverted...I would write:

$$\displaystyle V=\frac{m}{\rho}=\frac{2.80\text{ kg}}{8000\,\dfrac{\text{kg}}{\text{m}^3}}=\frac{7}{20000}\text{ m}^3$$

Okay, now we need to convert the 20.0 L of gasoline into m³. I would use the fact that 1 L is 1000 cm³...

#### Cbarker1

##### Active member
I think you have the volume inverted...I would write:

$$\displaystyle V=\frac{m}{\rho}=\frac{2.80\text{ kg}}{8000\,\dfrac{\text{kg}}{\text{m}^3}}=\frac{7}{20000}\text{ m}^3$$

Okay, now we need to convert the 20.0 L of gasoline into m³. I would use the fact that 1 L is 1000 cm³...
The answer for L to m^3 is .02 m^3.

#### MarkFL

Staff member
The answer for L to m^3 is .02 m^3.
$$\displaystyle 1\text{ L}=1\text{ L}\cdot\frac{1000\text{ cm}^3}{1\text{ L}}\cdot\left(\frac{1\text{ m}}{100\text{ cm}}\right)^3=\frac{1}{1000}\text{ m}^3$$

And so:

$$\displaystyle 20\text{ L}=20\cdot\frac{1}{1000}\text{ m}^3=\frac{1}{50}\text{ m}^3\quad\checkmark$$

Okay, so then what is the total volume of the can and gasoline when the can is full?

#### Cbarker1

##### Active member
$$\displaystyle 1\text{ L}=1\text{ L}\cdot\frac{1000\text{ cm}^3}{1\text{ L}}\cdot\left(\frac{1\text{ m}}{100\text{ cm}}\right)^3=\frac{1}{1000}\text{ m}^3$$

And so:

$$\displaystyle 20\text{ L}=20\cdot\frac{1}{1000}\text{ m}^3=\frac{1}{50}\text{ m}^3\quad\checkmark$$

Okay, so then what is the total volume of the can and gasoline when the can is full?
The Total volume is .02035 m^3.

#### MarkFL

Staff member
The Total volume is .02035 m^3.
Yes, although I would write:

$$\displaystyle V=\frac{407}{20000}\text{ m}^3$$

Now, we need to find the mass of the 20.0 L = 1/50 m³ of gasoline...we know the volume, and we know the mass density, and we have a formula that relates volume, density and mass...

#### Cbarker1

##### Active member
Yes, although I would write:

$$\displaystyle V=\frac{407}{20000}\text{ m}^3$$

Now, we need to find the mass of the 20.0 L = 1/50 m³ of gasoline...we know the volume, and we know the mass density, and we have a formula that relates volume, density and mass...
$$\rho=m/V$$
$$680=m/1/50\implies 680=50m\implies 680/50=m\implies 13.6=m$$

#### MarkFL

Staff member
$$\rho=m/V$$
$$680=m/1/50\implies 680=50m\implies 680/50=m\implies 13.6=m$$
To find the mass of the gasoline, I would write

$$\displaystyle m_g=\rho V=\left(680\,\frac{\text{kg}}{\text{m}^3}\right)\left(\frac{1}{50}\text{ m}^3\right)=\frac{68}{5}\text{ kg}$$

So, the total mass of the gasoline and the steel can is:

$$\displaystyle m_T=m_g+m_s=\left(\frac{68}{5}+\frac{14}{5}\right)\text{ kg}=\frac{82}{5}\text{ kg}$$

Recall, we found earlier:

$$\displaystyle V_T=\frac{407}{20000}\text{ m}^3$$

And so, what is the total mass density of the can full of gasoline?

#### Cbarker1

##### Active member
To find the mass of the gasoline, I would write

$$\displaystyle m_g=\rho V=\left(680\,\frac{\text{kg}}{\text{m}^3}\right)\left(\frac{1}{50}\text{ m}^3\right)=\frac{68}{5}\text{ kg}$$

So, the total mass of the gasoline and the steel can is:

$$\displaystyle m_T=m_g+m_s=\left(\frac{68}{5}+\frac{14}{5}\right)\text{ kg}=\frac{82}{5}\text{ kg}$$

Recall, we found earlier:

$$\displaystyle V_T=\frac{407}{20000}\text{ m}^3$$

And so, what is the total mass density of the can full of gasoline?
$$\frac{m_T}{V_T}=\rho_T$$
$$\frac{\frac{82}{5}}{\frac{407}{20000}}=\rho$$
$$\rho=806$$