# Temperature decreases in adiabatic expansion and gas laws

So I understand that the temperature decreases when a gas expands adiabatically. This is because there is no gain of heat from the surroundings, so the kinetic energy of molecules decreases in doing work on the surroundings, resulting in decreased temperature and pressure. Pressure decreases because same number of molecules are vibrating with a lesser kinetic energy in a larger volume.

Now my question is, why should the temperature of the gas not increase, as the volume is increasing as well. Volume is directly proportional to temperature, because when the volume increases, the gas molecules easily vibrate more vigorously, because intermolecular forces easily capture the slow moving gas molecules. But you may say, that the gas taken is an ideal one, and no intermolecular forces operate. Why then, do we say that Charles's Law holds good only for an ideal gas? In Charle's law too, the temperature is directly proportional to temperature. But if there are no Intermolecular forces, how does this happen?

• Are you referring to a reversible adiabatic expansion or to a free expansion? Feb 18, 2021 at 8:09
• en.wikipedia.org/wiki/Adiabatic_process Feb 18, 2021 at 8:34
• Temperature is the kinetic energy of the molecules. You have correctly explained in the first paragraph that it decreases. Feb 18, 2021 at 15:57
• @GiorgioP, I mean work against a piston. So adiabatic expansion. Thanks for the comment :-) Feb 18, 2021 at 17:39

So I understand that the temperature decreases when a gas expands adiabatically. This is because there is no gain of heat from the surroundings, so the the kinetic energy of molecules decreases in doing work on the surroundings, resulting in decreased temperature and pressure.

If by expanding adiabatically, you mean the gas does boundary work on the surroundings as in a piston cylinder arrangement (as opposed to a free expansion in a vacuum), then yes there will be a corresponding decrease in internal energy. Per the first law, $$\Delta U=Q-W$$, $$Q=0$$, therefore $$\Delta U=-W$$. But the decrease in internal energy depends only on temperature in the case of an ideal gas.

Pressure decreases because same number of molecules are vibrating with a lesser kinetic energy in a larger volume.

The decrease in translational kinetic energy is due to a decrease in the average translational velocities of he molecules, not due to decreased "vibration".

In any case, better to say the pressure decreases because the number of collisions per unit time on the walls is less due to the increase in volume and decrease in velocity, and the change of momentum (force) is less due to the decrease in velocity.

Now my question is, why should the temperature of the gas not increase, as the volume is increasing as well. Volume is directly proportional to temperature,...

Assuming an ideal gas, volume is directly proportional to temperature only when the pressure is constant.

$$pV=nRT$$

$$V=\frac{nRT}{p}$$

And the pressure is not constant in an adiabatic expansion.

Why then, do we say that Charles's Law holds good only for an ideal gas? In Charle's law too, the temperature is directly proportional to temperature.

Charles law only applies when the pressure remains constant. And as you've already stated, the pressure decreases in an adiabatic expansion.

Can it be possible that all the variables change simultaneously? Or is it necessary that one of the variables should remain constant?

For an adiabatic process volume, temperature, and pressure all vary simultaneously, but according to specific relationships.

For an ideal gas pressure and volume vary simultaneously according to

$$pV^{k}=constant$$

where $$k=c_{p}/c_{v}$$

Substituting for $$p$$ or $$V$$ using the ideal gas equation, temperature and volume vary simultaneously according to

$$nRTV^{k-1}=constant$$

and temperature and pressure vary simultaneously according to

$$nRTp^{1-k}=constant$$

For other reversible processes, one gas property may be held constant. Pressure is constant for an isobaric process, temperature is constant for an isothermal process, and volume is constant for an isochoric process.

Hope this helps.

• Thanks a lot. I just had a little doubt. Can it be possible that all the variables change simultaneously? Or is it necessary that one of the variables should remain constant? Feb 18, 2021 at 17:02
• @NishaPrakash I have updated my answer to respond to your follow up question. Hope it helps. Feb 18, 2021 at 17:30
• Thank you so so much. You have helped me a lot. Just a final question. Can you walk me through the mechanism of adiabatic expansion? I understood using the equation that all three variables CAN simultaneously vary. But when it comes to visualisation, I am able to picture the gas expanding, thus having a greater freedom of movement, thus increasing the transnational kinetic energy. But some of the internal energy is lost due to work on the piston. Thus the temperature should be constant. Feb 18, 2021 at 17:48
• @NishaPrakash I will be happy to update my answer to walk you thru the adiabatic expansion if that makes the answer acceptable. But as to your statement " am able to picture the gas expanding, thus having a greater freedom of movement, thus increasing the transnational kinetic energy" what makes you think the "greater freedom of movement" increases the translational kinetic energy? Feb 18, 2021 at 17:55
• Please do correct me where I'm wrong..So in case of a real gas, it would be the Intermolecular forces. When a gas expands a lot(I'm not talking about intermediate expansion as then, intermolecular attractive forces would come into picture), the molecules come outside of the influence of the intermolecular forces exerted by neighboring molecules. But in case of an ideal gas, which has no Intermolecular forces, I just assumed a greater volume would mean greater freedom of movement Feb 19, 2021 at 3:03