Thermodynamics- energy, pressure, volume

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Homework Help Overview

The problem involves a thermodynamics scenario where the internal energy of a system increases due to heat absorption at constant pressure. Participants are tasked with determining the change in volume based on given energy values.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the isobaric equation and the first law of thermodynamics. There is confusion regarding the correct interpretation of work done on the system and how to relate it to the change in internal energy and heat absorbed.

Discussion Status

Several participants are engaged in clarifying the definitions of internal energy, heat, and work. Guidance has been provided regarding the relationship between these quantities, but there remains some confusion about the calculations and the correct values to use.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the relationships between the thermodynamic variables involved.

name_ask17
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Homework Statement


The internal energy of a system increases by 1350J when the system absorbs 1150J of heat energy at a constant pressure of 1.01E5 Pa. By how much does the volume of the system change?


Homework Equations


Im not sure how to approach this but would i use the isobaric equation since the pressure is constant?
w=-P(change in V)


The Attempt at a Solution


I attempted to take the change in the work (1350) and divide out pressure, but that did not give me the correct answer.
(The answer should be 2E-3m^3, but i just don't know how to get there)
 
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name_ask17 said:
I attempted to take the change in the work (1350)
That's not the work done on the system. (That's the total change in internal energy.)

(But your method is exactly right. Just find the correct value for the work and you'll be fine.)
 
Last edited:
Doc Al said:
That's not the work done on the system. (That's the total change in internal energy.)

(But your method is exactly right. Just find the correct value for the work and you'll be fine.)

But I am still confused on what i am supposed to use for work
 
name_ask17 said:
But I am still confused on what i am supposed to use for work
What does the 1st law of thermo tell you?
 
U=q+w
 
is that correct? and is so, how do i use that in this problem? i am confused.
 
name_ask17 said:
U=q+w
Exactly. You're given U and q, so what's w?
 
the answer to this problem should be 2E-3 but i can't seem to get that
 
i just add them? but then i get the answr of .023. and that's not right
 
  • #10
name_ask17 said:
i just add them?
Add what?

U = ?
Q = ?

Solve for W.
 
  • #11
can you explain to me what i am doing wrong?
 
  • #12
which one is U? would it be the 1350?
 
  • #13
or would that be the q?
 
  • #14
name_ask17 said:
which one is U? would it be the 1350?
Yes, U = change in internal energy.
 
  • #15
so does that make the w 200?
 
  • #16
name_ask17 said:
so does that make the w 200?
Yes.
 
  • #17
thank you!
 

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