Temperature in an isobaric process

Question

I have a certain conceptual issue - I'm solving heat engine problems and I found something difficult to understand. Let's take an isobaric part of a cycle of an engine, let's say that the ideal gas is compressed from $2V$ to $V$. How does the temperature change? From the Clapeyron equation in the earlier point it is $T_2 = 2Vp/nR$ and in the latter point $T_1 = Vp/nR$ (it's labeled "1", because the cycle closes there).

This would mean that compressed gas has lower temperature that the uncompressed one? This is false isn't it? Could someone please explain this to me?

Answer 1

If you rearrange the ideal gas law to be expressed in terms of pressure $$ P=\frac{NRT}{V}\qquad\Rightarrow\qquad \frac{T_1}{V_1}=\frac{T_2}{V_2} $$ where the right hand equation assumes it is an isobaric process with no mass exchange. So, in an isobaric process temperature and volume vary inversely. If the volume decreases then the temperature must go up. Your confusion probably arises because we don't cool gasses by compressing them at a fixed pressure. The pressure, in fact, rises dramatically as we compress them. If you allow pressure to vary, then the above equality can be written as $$ \frac{T_1}{P_1V_1}=\frac{T_2}{P_2V_2} $$ so as long as the pressure rise is more dramatic than the volume reduction the temperature goes down.

Answer 2

Yes, Temperature is halved.

You are keeping the pressure constant. You are just imagining pressing a piston in which pressure is not constant. Volume is reducing. But we have to keep the pressure same. So we must reduce kinetic energy of our particles to keep the pressure same. The rest is obvious.