dani

I found the graph above of an ideal gas's transformation in a book, and it says that between 1-2 the heat (Q) is positive and at point 2 it is 0, which I understand because it's an adiabatic intersecting that and after that it becomes negative. Why is it first bigger than 0 and then smaller?

Attached Thumbnails

 

Woody

The key is the the slope of the pressure to volume curve relative to the (local) adiabatic curve.

__________________
~\o/~

dani

Can you explain what exactly does it mean if the slope of the adiabatic is steeper on the upper side?

Woody

For an "ideal" gas at constant temperature,
pressure and volume are inversely proportional to each other.
This describes the Adiabatic curve on your graph.

If the variation of Pressure relative to Volume is following the adiabatic curve, then it is will remain at a constant temperature.
If the variation of Pressure relative to Volume departs from the adiabatic curve, (as it does in your chart), then the temperature will change.

If the rate of change of Pressure with respect to Volume is faster than the adiabatic limit it will produce heat.
If the rate of change of Pressure with respect to Volume is lower than the adiabatic limit it will get colder.

see also this link <Wikipedea:Ideal Gas Law>

__________________
~\o/~

studiot

Dani, can you tell us what the book is and if it is in English please?

I am having trouble understanding the graph you show.

The straight line is not the P_V curve for an ideal gas, and gases don't 'transform'.

The plot of pressure against the reciprocal of volume, 1/V is a straight line.

An ideal gas at point 1 on the straight line will not follow it to point 2 as the gas expands.