Thermodynamics -- Internal Energy

[BurningUrge]

I've read back and forth in my chapter and tried consulting with my forumla sheet, but I cannot seem to find the correct formula. I bet I am looking at this question wrong, but I am fairly sure I need to use the -100 kJ for something to find the final volume and pressure , along with the given pv²=constant.

Basically, given the information below, how would you go about starting to solve this?

Homework Statement

A mass of gas at an initial pressure of 28 bar, and with an internal energy og 1500 kJ, is contained in a well-insulated cylinder of volume 0.06m³. The gas is allowed to expand behind a piston until it's internal energy is 1400 kJ; the law of expansion is pv²=constant. Calculate:

1) the work done (Already calculated, -100kJ)
2) the final volume
3) the final pressure

Homework Equations

The Attempt at a Solution

 

[Chestermiller]

Please show your calculation of the work done, specifying whether it is the work done by the gas on the surroundings, or the work done by the surroundings on the gas.

Chet

 

[BurningUrge]

Okay, so the work done is by the internal energy as it expands.

So Q + W = U₁-U₂. Since there is no heat transfer, only work done, it's basically only W = U₁-U₂.

W = 1500 kJ - 1400 kJ => W = 100 kJ. But since it's an expansion done BY the internal energy as it is allowed to expand, it's -100kJ.

 

[Chestermiller]

Okay, so the work done is by the internal energy as it expands.

So Q + W = U₁-U₂. Since there is no heat transfer, only work done, it's basically only W = U₁-U₂.

W = 1500 kJ - 1400 kJ => W = 100 kJ. But since it's an expansion done BY the internal energy as it is allowed to expand, it's -100kJ.

The equation should be Q + W = U₂-U₁. That's why W is negative (using the sign convention that W represents work done on the system).

Now you need to develop another equation for expressing the work in terms of the pressure and volumes. What is the equation you learned for determining the work as an integral involving pressure and volume?

Chet

 

[BurningUrge]

That would quite probably be this nifty little thing;

W = -∫pdV , from position 1 to 2. (Couldn't find the icon for definite integral)

Since pv² = Constant, I rewrite the equation to -> -∫ c/V² dV, => -c ∫ 1/V² dV => -p₁V₁² ∫ 1/V dV.

When I can't write the correct definite integral it looks abit messy, but the finished integration should look like this;

-p₁V₁² (-1/V₂-(-1/V₁)

Finishing this off with :

-p₁V₁² (-1/V₂-(-1/V₁) = -100kJ

Am I currently on the right path?

 

[Chestermiller]

That would quite probably be this nifty little thing;

W = -∫pdV , from position 1 to 2. (Couldn't find the icon for definite integral)

Since pv² = Constant, I rewrite the equation to -> -∫ c/V² dV, => -c ∫ 1/V² dV => -p₁V₁² ∫ 1/V dV.

When I can't write the correct definite integral it looks abit messy, but the finished integration should look like this;

-p₁V₁² (-1/V₂-(-1/V₁)

Finishing this off with :

-p₁V₁² (-1/V₂-(-1/V₁) = -100kJAm I currently on the right path?

Yes. I haven't checked your "arithmetic," but you're on the right path.

Chet

 

[BurningUrge]

Okay, so I did that. And I came up with the right answers.

V₂=0.148m³

And using that I could find the pressure as well with p₁V₁²=p₂V₂²

Ending at p₂=4.6 bar.

Greatly appreciate your time and help!

 

[Kc emefo]

That would quite probably be this nifty little thing;

W = -∫pdV , from position 1 to 2. (Couldn't find the icon for definite integral)

Since pv² = Constant, I rewrite the equation to -> -∫ c/V² dV, => -c ∫ 1/V² dV => -p₁V₁² ∫ 1/V dV.

When I can't write the correct definite integral it looks abit messy, but the finished integration should look like this;

-p₁V₁² (-1/V₂-(-1/V₁)

Finishing this off with :

-p₁V₁² (-1/V₂-(-1/V₁) = -100kJ

Am I currently on the right path?

I tried substituting into this formula you used but am not getting V2=0.148,am getting different answer.can you show your working here on how you got V2= 0.148 given the intial pressure and volume

 

[berkeman]

Welcome to the PF.

I tried substituting into this formula you used but am not getting V2=0.148,am getting different answer.can you show your working here on how you got V2= 0.148 given the intial pressure and volume

This thread has been inactive for 5 years, but perhaps @Chestermiller can offer his thoughts...

 

[Kc emefo]

Welcome to the PF.

This thread has been inactive for 5 years, but perhaps @Chestermiller can offer his thoughts...

Thank you,can you help me out with the appropriate substitution to arrive at the given answer V2=0.148m @berkeman

 

[Chestermiller]

Thank you,can you help me out with the appropriate substitution to arrive at the given answer V2=0.148m @berkeman

Please show us the details of what you have done so far.

 

[Kc emefo]

Substituting with the value of P1=28bar V1=0.06 and n=2 into the given equation:
2.8× 10^6×0.06^2(1/V2-1/0.06)=-100
Opening bracket:
10080(1/V2-1/0.06)=-100
10080/V2-1/0.06=-100
10080/V2-16.66=-100
Cross multiple and make V2 subject of formula:
10080-16.66=-100V2
10063.34/-100=-100V2/-100
V2=-100.6334
That is what am getting, please correct me if am wrong

 

[Chestermiller]

You are working in kJ and kbars. So that 2.8x10^6 should be 2.8x10^3.

 

[Kc emefo]

You are working in kJ and kbars. So that 2.8x10^6 should be 2.8x10^3.

Oh I got that,jus substituted 2.8×10^3 instead of 2.8×10^6 and my V2=1.5792 which is wrong,am still not getting the suppose as V2=0.148

 

[Chestermiller]

Oh I got that,jus substituted 2.8×10^3 instead of 2.8×10^6 and my V2=1.5792 which is wrong,am still not getting the suppose as V2=0.148

check your math

 

[Kc emefo]

check your math

Oh thank you very much sir,I just figured out my mistake and corrected it
Am grateful for that correction