Adiabatic processes and the Ideal Gas Law in my first year chemistry textbook it wrights that in adiabatic proccess when there is a system of canister filled with gas and a piston above(and wheights on it) if we let the piston move upward by removing some of the weights the temperature of the gas will decrease since the gas has made work. 
i understand that in the opposite case when we compress a gas in adiabatic proccess the temperature will increase.

**my question is: isn't it contradicts the ideal gas law which says that the volume of the gas is proportional to the temperature and vice versa

PV=nRT

so when the volume of the gas is increasing, the temperature should increase.

also by intuition: if the volume is increased, there is less pressure , the molecules has now more space and theire kinnetic energy should increase and therefore also the temperature.

how can i look at this?

thanks
 **
No, it doesn't contradict the ideal gas law at all. When you are saying that rise in volume means increase of temperature, you are inherently assuming that Pressure remains constant. But that is not the case at all in an adiabetic expansion.

In fact, you should realise that it is the product $PV$ which is decreasing. So, Volume might increase, but pressure drops even faster than volume, that's why the product $PV$ overall decreases. If some of the weights are removed, the gas expands slowly according to the equation, $PV^{\gamma} = C$. For most of the gases $\gamma &gt; 1  .(\gamma = \frac{C_p}{C_V})$ Thus by simple observation, you can see if Volume increases $x$ times, Pressure will decrease $x^{\gamma}$ times, which is more dominating in the product.Thus, the product $PV$ will definitely decrease with increase in volume in case of an adiabetic process. 

Also, your second argument is also not acceptable, because you already realise that pressure will decrease and by Kinetic theory of ideal gases $PV = \frac{1}{3}m\ n\ v^2_{rms}$. Thus if you understand the decrease in the product $PV$, you will also understand that $v^2_{rms}$ will also decrease, and hence the kinetic energy of gas molecules also. 

Also you are thinking in a much more complicated way towards the situation. Think simply. For ideal gases, internal energy ($U$) is a measure of the translational kinetic energy of gas molecules and also $U = \frac{3}{2}RT$ depicting that for ideal gases, kinetic energy is only a function of temperature only. Also, by the first law of thermodynamics, $dU = \delta q + \delta W$ ( $\delta $ here represents inexact differentials). For adiabetic process, $\delta q =0$. So, $dU = \delta W$. Now, $\delta W$ is work done on the system which is negative for an adiabetic expansion, which makes $dU$ also negative, i.e. internal energy decreases and so does temperature.