Find work done on an isobaric system

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Homework Help Overview

The problem involves calculating the work done on a 1kg box of Aluminum during an isobaric process as its temperature increases from 22°C to 40°C at a constant pressure of 1 atm. The context is thermodynamics, specifically focusing on work done in relation to volume change due to thermal expansion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for work done in an isobaric process and question the change in volume. There are attempts to clarify the correct expression for work by considering the initial and final volumes.

Discussion Status

Some participants have provided guidance on the correct formulation of the work done, highlighting the need to account for the initial volume in the calculations. There is an acknowledgment of a potential oversight in the original attempt, but no consensus on the final answer has been reached.

Contextual Notes

Participants note that the original poster's calculations yield a result that differs from the expected answer, indicating possible miscalculations or assumptions regarding the volume change. The discussion reflects on the importance of accurately applying the principles of thermal expansion and work in thermodynamic systems.

Feodalherren
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Homework Statement


A 1kg box of Aluminum initially at 22C is warmed at P=1atm so that its T increases to 40C. Find
a) The work done on the system


Homework Equations



W= -∫PdV

The Attempt at a Solution



The pressure is constant at 101,300Pa

D=m/V

So V=m/D

D of Al at room temp = 2700kg/m^3

Therefore

[itex]W= -101,300Pa \frac{1kg}{2700kg/m^{3}}(1+24x10^{-6}(18))^{3}[/itex]

Where ΔT=18 and 24x10^-6 is the linear coefficient of expansion of Aluminum.

= -37.6J

Not the right answer, it should be around 48mJ. I can't see what I'm doing wrong here.
 
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Work is PΔV. What is the change of volume?

ehild
 
It should be

[itex]W= -101,300Pa \frac{1kg}{2700kg/m^{3}}((1+24\times 10^{-6}(18))^{3}-1)[/itex]

because you forgot to subtract the initial volume. Essentially the same result, which retains only the linear term, is:

[itex]W= -101,300Pa \frac{1kg}{2700kg/m^{3}}(3)(24\times 10^{-6}(18))[/itex]

Chet
 
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Oh yeah I totally missed that. Thanks.
 

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