First law of a thermodynamic system

[maitake91]

Homework Statement

Heat is transferred through the wall of a vertical cylinder to air trapped inside it by a frictionless piston of mass 10kg and radius 0.05m. When 300J of heat is transferred to the air the piston rises by 0.2m against the atmospheric pressure of 100kPa. Determine the change in internal energy of the trapped air when 300J is transferred and the piston rises 0.2m.

Relevant Equations

(Qin - Qout)+(Win - Wout)+(Emassin - Emassout) = ∆U+∆KE+∆PE

After crossing out all the variables which I think equals 0 in the equation, I was left with:
∆PE + ∆U = Qin
mg (0.2) + ∆U = 300
10*9.81*0.2 - 300 = - ∆U
= -280J

This was the answer I derived. However, the correct answer was supposed to be 123.3J. Please can someone explain to me how to get the correct answer and where I did wrong? This may not be relevant but the pressure of the trapped air was worked out in earlier questions, which equals to 11240Pa.

 

[Chestermiller]

The potential energy of your "system" (the air) does not change. The mg(0.2) is part of the work that the air does on its surroundings. Also, as Kyouran points out, you omitted the additional work the air does in pushing back the surrounding atmosphere.

 

[Kyouran]

I deleted my previous post to try and make it more helpful. Sometimes it helps to break up the system into 2 parts. In this case, consider the energy balance of the piston alone and the energy balance of the gas (without the piston alone). Note that the gas does work onto the piston (what is this equal to?) and in addition, the gas also performs work as its volume expands (volume expansion work).

 

[maitake91]

I deleted my previous post to try and make it more helpful. Sometimes it helps to break up the system into 2 parts. In this case, consider the energy balance of the piston alone and the energy balance of the gas (without the piston alone). Note that the gas does work onto the piston (what is this equal to?) and in addition, the gas also performs work as its volume expands (volume expansion work).

Thank you for your kind explanation, I was able to get it right this time! XD

 

[maitake91]

The potential energy of your "system" (the air) does not change. The mg(0.2) is part of the work that the air does on its surroundings. Also, as Kyouran points out, you omitted the additional work the air does in pushing back the surrounding atmosphere.

Thank you for you help, that cleared a lot things up!