Adding energy to an ideal gas without changing internal energy

1. May 7, 2014

Numaholic

Hi everyone.

I thought of a question which has been bothering me. It is: Is there a case where energy is added to an ideal gas of constant amount where the energy added will equal the work done by the gas.

My thoughts: The energy of an ideal gas is proportional to the temperature. If the gas is to do work ΔW≠0 W=Fx=PV. If the gas is to do work on the environment then PV will increase. Using the ideal gas equation P1V1/T1=P2V2/T2. If P2V2 > P1V1 then T2 > T1. Therefore energy of the system has increased and it is impossible for the energy added to the ideal gas to equal the work done by the gas.

I don't know if I made an incorrect assumption or faulty argument, but the result doesn't seem intuitive to me. Any thoughts would be appreciated!

2. May 7, 2014

Staff: Mentor

The amount of work is ∫PdV, not Δ(PV). This integral does not have to be equal to zero when the temperature is held constant (and PV is constant). For an isothermal reversible expansion, ∫PdV = nRT ln(Vfinal/Vinitial)

Chet

3. May 7, 2014

Simon Bridge

You are looking for a situation where $Q = W$

Your argument shows that the work done by an ideal gas is a function of the temperature change.
It has not shown that the heat added to the system cannot be as high as the work done by the system.

You can try seeing what the laws of thermodynamics have to say about Q=W ... or just have another look at the different thermodynamic processes you know about.
You have a nice hint in your working - it loks like any process where the temperature change is non-zero won't have Q=W doesn't it. So look for a process where T2=T1.

[edit: too slow - thanks chet]