1st Law of Thermo, work in a piston.

AI Thread Summary
The discussion focuses on calculating the total work done during the expansion of air in a piston-cylinder system under two different conditions. Initially, work is calculated for a constant pressure expansion from 0.1 m³ to 0.3 m³, yielding 40,000 J. For the second expansion at constant temperature, the user struggles with the variable pressure and seeks to use the ideal gas law to express work. A suggestion is made to simplify the calculation by using the initial pressure and volume instead of finding the new pressure. The final computed work for the second expansion is approximately 30,649.54 J, confirming the approach is valid.
twmggc
Messages
4
Reaction score
0

Homework Statement


Air is expanded in a piston-cylinder arrangement at a constant P of 200kPa from a volume of 0.1 m3 to a volume of 0.3m3. Then the temperature is held constant during an expansion to a volume of 0.5m3. Predict the total work done in the air.


Homework Equations


W = ∫PdV , PV = nRT


The Attempt at a Solution


For the first part of the expansion I used:
W = ∫PdV = P∫dV = P(V2 - V1) = 200,000Pa ( 0.3-0.1)m3
and got W = 40,000J.

The next expansion is what is confusing me.
Since Pressure is no longer constant I need to leave it in the W = ∫PdV equation.
So, I transform this equation using the ideal gas eqn. PV = nRT and get:
W = nRT∫dV/V = nRT*ln(V3/V2)
But, without knowing the n or the T how do I get the work?

Thanks!
 
Physics news on Phys.org
Trick: Replace the nRT in your expression for W with some other expression (using the ideal gas law).
 
Since Temperature is constant for this part I can assume P2V2=P3V3 and find P3 and then replace nRT in my work equation with P3V3 (PV = nRT).
After my computations I got 30,649.54J which sounds within reason.

Was this what you were thinking?
 
Yes. But there's no need to find P3. You can use P2 and V2 instead of P3 and V3.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top