[StillLearning]

The question:

The air within a piston equipped with a cylinder absorbs 565 j of heat and expands from an initial volume of 0.10 L to a final volume of 0.84 L against an external pressure of 1.0 atm.  What is the change in internal energy of the air within the piston?


My approach:

Using the formula -w=P*volume_final-volume_initial I get w=-.74 j.

Then, using the formula Delta E=q+w I get the following: Delta E=565j+(-.74 j), thus the Delta E=564.26j. 

However, this answer seems to be incorrect.  Can anyone tell me where i'm going wrong?  Thanks.

[Astrokel]

always check your units, it should be pascal.

[StillLearning]

always check your units, it should be pascal.

You're saying the atm should be converted to pascal?  Isnt the formula w=-P*(pressure_final-pressure_initial) with the pressure measured in atmospheres?  I believe that's what my book says.  Maybe i'm misunderstanding you. 

[Astrokel]

Hey yes it needs to be covert from atm to pascal.

w=P*(Volume_final - Volume_initial)

where P  = Pressure in Pascal, V = Volume in m³ or L

of course work done is scalar so you don't need to put the negative sign infront, but when in calculating internal energy change, you subtract because it is work done by the system.

[Borek]

Isnt the formula w=-P*(pressure_final-pressure_initial) with the pressure measured in atmospheres?

Doesn't matter what units pressure is masured in, as long as you are using constistent units throughout all formula.

Different example, but one that you should easily grasp:

s = v t

Distance covered equals product of time and speed. Does it matter if you measure speed in km/h or mph, or m/s or even feet per second? No, formula holds in all cases, just time have to be expressed in second or hours and calculated distance will be in feets, meters, kilometers or miles - but that's just units, and formula is universal.

Same holds for EVERY correct physical formula you will ever encounter.

[StillLearning]

So my units are consistent, right?  If not, where did I go wrong?  I think I understand what both of you are saying, but I still dont see where my approach to the problem is incorrect.  Would either of you mind working the problem so I can see the correct way to solve it?  Thanks in advance. 

[Astrokel]

like Borek has mention, check your units consistently and you won't go wrong

work done = p :delta:V

[workdone] = Kg m² s^-2
[delta V] = m³

[P] =   Kg m² s^-2
        ------------   = Kg m^-1 s^-2
           m³

Did you realize your units of P, Kg m^-1 s^-2 is actually equal to Pascal, so you should change your pressure to Pa!