MHB The average density of the gasoline in the steel gas can is 806 kg/m^3.

AI Thread Summary
The average density of a full steel gas can, which holds 20 liters of gasoline, is calculated to be 806 kg/m³. The mass of the steel can is 2.80 kg, and the density of gasoline is given as 680 kg/m³. To find the total density, the volume of the gasoline is converted from liters to cubic meters, resulting in 0.02 m³. The total mass of the gasoline is calculated to be 13.6 kg, leading to a combined mass of the gasoline and can of 16.4 kg. The final density is derived from the total mass divided by the total volume, confirming the average density of 806 kg/m³.
cbarker1
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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
 
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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?
 
MarkFL said:
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.
 
Cbarker1 said:
the volume is 2857.14 m^3.

I think you have the volume inverted...I would write:

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

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...
 
MarkFL said:
I think you have the volume inverted...I would write:

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

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.
 
Cbarker1 said:
The answer for L to m^3 is .02 m^3.

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

And so:

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

Okay, so then what is the total volume of the can and gasoline when the can is full?
 
MarkFL said:
$$1\text{ L}=1\text{ L}\cdot\frac{1000\text{ cm}^3}{1\text{ L}}\cdot\left(\frac{1\text{ m}}{100\text{ cm}}\right)^3=\frac{1}{1000}\text{ m}^3$$

And so:

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

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.
 
Cbarker1 said:
The Total volume is .02035 m^3.

Yes, although I would write:

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

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...
 
MarkFL said:
Yes, although I would write:

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

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...

$$\rho=m/V$$
$$680=m/1/50\implies 680=50m\implies 680/50=m\implies 13.6=m$$
 
  • #10
Cbarker1 said:
$$\rho=m/V$$
$$680=m/1/50\implies 680=50m\implies 680/50=m\implies 13.6=m$$

To find the mass of the gasoline, I would write

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

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

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

Recall, we found earlier:

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

And so, what is the total mass density of the can full of gasoline?
 
  • #11
MarkFL said:
To find the mass of the gasoline, I would write

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

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

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

Recall, we found earlier:

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

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

$$\frac{m_T}{V_T}=\rho_T$$
$$\frac{\frac{82}{5}}{\frac{407}{20000}}=\rho$$
$$\rho=806$$
 
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