Isothermal Ideal gas Thermodynamics

Isothermal Process: ΔT=0 therefore ΔE=0. 1st law of thermodynamics states ΔQ(in)=ΔE-W(on), thus here ΔQ(in)=W(on). However, why isn't the following true? ΔQ=mcΔT, ΔT=0 therefore ΔQ=0.
- Youssef Hsheimi (age 22)
Montreal, QC, Canada

Fisrt, I should point out that the assumption "ΔT=0 therefore ΔE=0" applies only to an ideal classical gas. In general, E will depend not only on T but also on volume V. So let's assume that your question only is about ideal classical gases.

Then you ask "why isn't the following true? ΔQ=mcΔT, ΔT=0 therefore ΔQ=0...." Presumably here "m" is the mass and "c" is the heat capacity (at constant V? constant p?) per unit mass. At any rate,  "ΔQ=mcΔT" is simply not true for most processes in which V changes.  Why would it be?

Another point to watch out for, although not needed here, is that the specific heat "c" generally depends on T, even for ideal classical gases. 

Mike W.

(published on 11/06/2014)