Compression of air in closed tank

[trousseli]

Hello,

 Here's my problem:

 I have a firm steel rectangle.

 dimensions :
 Length: 530mm
 Height: 610mm
 Width: 15mm

 This one is filled with oil up to 550 mm high. The rest of the volume is occupied by air. All this under normal pressure and temperature conditions.
 Now I put this steel rectangle on a heating device that will heat the oil up to 85C. Taking into consideration the thermal expansion of the oil, its volume will increase by 'x' m3 due to the increase in temperature (the thermal expansion of the steel will be neglected).

 My question is this:

 The new volume of oil will affect the air pressure as it will be compressed. What will be the value of this air pressure?

I found the volume expansion of oil but not sure if it is correct. Could anyone enlighten me on this please?

 Thank you for your feedback,

 Friendly

[studiot]

Your physical analyis is correct (I haven't checked the arithmetic) provided that beta is the volume coefficient of expansion, not the linear coefficient.

https://opentextbc.ca/physicstestbook2/chapter/thermal-expansion-of-solids-and-liquids/

Also provided that it is permissible to ignore the expansion of the steel box (steel has a significant expansion coefficient).

I agree that the oil can be regarded as incompressible so the air 'takes up' all the volume loss due to expansion.

So now you can calculate the volume of air by subtraction in both the cold and heated conditions.

Then you need Boyles law

Pressure_original x Volume_original = Pressure_compressed  x Volume_compressed

It is probably easiest to work in Bars which makes the expanded pressure equal to

Pressure_compressed = Volume_original / Volume_compressed

since you started with 1 bar of pressure.

 

Does this help?

[trousseli]

Hi Studiot,

Thanks for your reply. 

I have just made the calculation using the Boyles low and I have found a compressed air pressure of 1.32 bar (Volume_original / Volume_compressed =0.477L/0.36L).

However, I made an other calculation by using the PV=NRT formula (Since we talk about air, I assume using PV=NRT is allowed) and got an other result for the compressed air pressure (1.62 bar).

You can see how I got this that result by looking at the file attached.

[Bufofrog]

2 hours ago, trousseli said:

Hello,

 Here's my problem:

 I have a firm steel rectangle.

 dimensions :
 Length: 530mm
 Height: 610mm
 Width: 15mm

 This one is filled with oil up to 550 mm high. The rest of the volume is occupied by air. All this under normal pressure and temperature conditions.
 Now I put this steel rectangle on a heating device that will heat the oil up to 85C. Taking into consideration the thermal expansion of the oil, its volume will increase by 'x' m3 due to the increase in temperature (the thermal expansion of the steel will be neglected).

 My question is this:

 The new volume of oil will affect the air pressure as it will be compressed. What will be the value of this air pressure?

I found the volume expansion of oil but not sure if it is correct. Could anyone enlighten me on this please?

 Thank you for your feedback,

 Friendly

Do not forget to also take into account the increase in the air pressure due to heating the air.

In other words you have 2 effects going on.  You have the increase in pressure due to the decreased volume + the increase in air pressure due to the increased temperature.

The relationship PV=nRT can be used for both of these effects.  Good luck with your homework...   

[trousseli]

Hi Bufrofrog,

I actually did use the PV=nRT relationship and submitted above. I took into account the new volume of air decreased by the oil expansion as well as the hot temperature of 358.15K (85C). Does it seem correct to you?

Thanks

[studiot]

1 hour ago, Bufofrog said:

Do not forget to also take into account the increase in the air pressure due to heating the air.

No this is wrong.

 

1 hour ago, trousseli said:

However, I made an other calculation by using the PV=NRT formula (Since we talk about air, I assume using PV=NRT is allowed) and got an other result for the compressed air pressure (1.62 bar).

 

What did you use for N?

[trousseli]

First, I calculated n with the following input values:

V --> Initial volume of air (0.477L=0.000477m3)

P--> Initial air pressure (1 Bar=100000 Pa)

T--> Initial temperature (293.15K)

n=PV/RT

n=(100000*0.000477)/(8.31*293.15)=0.01958 moles

I used the same n value for the Air compressed pressure as the amount of substance does not change.

The only variables that changed were the volume (0.000360473m3 instead of 0.000477m3 because of the oil expansion) and the temperature (358.15K instead of 293.15K).

Thanks

 

 

[studiot]

Hey you are correct, I was wrong, but you don't need to go to the trouble of calculating N.

you can use

[math]\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}[/math]

to obtain

[math]{P_2} = \frac{{{T_2}{P_1}{V_2}}}{{{T_1}{V_2}}}[/math]

directly.

Well done both of you, +1 each.

Edited February 26, 2019 by studiot

[trousseli]

That's great, thank you so much for your help!

[Bufofrog]

3 hours ago, studiot said:

No this is wrong.

Why?  

Edited:  Are you saying the temperature increase can be taken into account in the same equation, if so I agree.  My point is 2 things are happening to increase the pressure.

Edited February 26, 2019 by Bufofrog

[studiot]

37 minutes ago, Bufofrog said:

Why?  

Edited:  Are you saying the temperature increase can be taken into account in the same equation, if so I agree.  My point is 2 things are happening to increase the pressure.

I realised that I was wrong and corrected it in the rest of this post

3 hours ago, studiot said:

Well done both of you, +1 each.

 

[Bufofrog]

4 hours ago, studiot said:

I realised that I was wrong and corrected it in the rest of this post

 

Good deal.