I've read that adibatic expansion always results in a decrease in temperature.

Since temperature is average kinetic energy, I don't get how kinetic energy can decrease when we increase the volume.

Kindly consider an example. Close the mouth of below syringe, when it is at 1ml, with your index finger, and then pull the syringe to 10ml. The volume of the gas has increased from 1ml to 10ml. Why would the average kinetic energy decrease in this case ?

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If the walls of the syringe are not moving then the collisions between the air molecules and the walls can be considered to be elastic.
This means that on average the kinetic energy of the molecules before collision with the walls is the same as the kinetic energy of the molecules after collision with the walls.

Now move one of the walls (the piston of the syringe) outwards.
In this case the speed and hence the kinetic energy of the molecules after rebound with the outwardly moving wall is going to be less than the speed and kinetic energy before they hit the moving wall.
So the average kinetic energy of the air molecules in the syringe has decreased which means that the temperature of the air has decreased.

  • $\begingroup$ If I understand correctly, when I start pulling the syringe, molecules adjacent to the moving syringe wall rebound with a decreased kinetic energy(because the wall is moving away from the molecules.) This manifests as a decrease in the average kinetic energy. OMG you just blew my mind! Thank you XD $\endgroup$ – AgentS Sep 3 '18 at 15:38

Consider the first law of thermodynamics

$$ \Delta U = W + Q,$$ where $U$ is the internal energy, $W$ is the external work, and $Q$ is heat transferred. In an adiabatic process, we have $Q=0$ by definition.

Then we have $$ dU = W = -PdV.$$

In the case of opening the syringe, the system of the gas has a volume increase (from 1 mL to 10 mL). That means that $dV > 0$, and thus the entire term is $-PdV$ is negative. So

$$ \Delta U < 0.$$

In this gas we see a loss of internal energy. By the equipartition theorem

$$ U = \frac{1}{2}fNkT,$$

internal energy is proportional to temperature, so the gas' temperature must also decrease. This is why the kinetic energy decreases; kinetic energy is related to temperature by the root-mean-square speed

$$ v_{\rm rms} = \sqrt{\frac{3kT}{m}}.$$

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    $\begingroup$ How do you see a loss of internal energy if the gas does no work (it does not cause it to expand) and no work is done on the gas (the external force does not cause it to compress). $\endgroup$ – Bob D Sep 3 '18 at 14:44
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    $\begingroup$ Work is done on the gas. When you open the syringe, you change its volume and do negative work on the system. To conserve energy, the system's internal energy must correspondingly decrease. $\endgroup$ – zhutchens1 Sep 3 '18 at 14:52
  • $\begingroup$ Oh right, right. If I compress the syringe, I am indeed doing work on the gas(adiabatic compression). This should increase the temperature of the gas. Similarly if I pull the syringe, am I doing negative work on the gas ? $\endgroup$ – AgentS Sep 3 '18 at 14:56
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    $\begingroup$ @zhutchens1.Ok I agree. Work is being done by an external agent. It has to be accounted for somehow. Normally when work is done on the gas it involves compression. When work is done by the gas it normally involves expansion. However, on rethinking, we can view the force pulling on the syringe as one that reduces the outside pressure, allowing the gas to do work and expand. I have deleted my answer. $\endgroup$ – Bob D Sep 3 '18 at 15:08
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    $\begingroup$ There are 3 forces acting on the piston during the expansion: your pulling force, the pressure of the outside atmosphere (acting in the opposite direction), and the (lower) pressure of the gas (acting in the same direction as the piston motion and your pulling force). Most of the work you do involves moving the outside atmosphere back. This is much more than the work the gas does. But the gas does do some work at the inside face of the piston, equal to its pressure times the piston area times the displacement of the piston. $\endgroup$ – Chet Miller Sep 3 '18 at 15:44

It is like throwing a ball in the air. As the ball gets distant from the earth, it decelerates. This is analogous to why an increase in volume decreases temperature. The interactions between gas molecules and between gas molecules and walls play the role of gravity.


This answer is intended to support and augment the answer given by zhutchens1 by showing how to calculate the reduction in temperature.

When you pull on the syringe you are doing the equivalent of reducing the outside pressure. The differential allows the gas to expand and do work at the expense of its internal energy (average translational kinetic energy of the molecules). So yes, the temperature will decrease. With knowledge of the properties of the gas, measurement of the force you are exerting to hold the syringe at 10 ml, and the following stipulations, you could calculate the change in temperature.

Lets say you pull on the syringe very slowly so that the adiabatic expansion is considered reversible (isentropic). Let’s further assume that the gas is an ideal gas.

From 1st law (see comments below about the form of this law)

$$\Delta u = q – w$$

For an adiabatic process, $q=0$ so

$$\Delta u = -w$$

The work done by a reversible adiabatic (isentropic) process for an ideal gas is given by

$$w =\frac{P_2v_2 – P_1v_1}{1-k}$$ where

Also, for an ideal gas, any process: $$\Delta u = c_v\Delta T$$

Putting the above together we have:

$$\Delta T = -\frac{(P_2v_2 – P_1v_1)}{c_v(1-k)}$$

Initial conditions

$$P_1 = 1 atm$$ $$v_1 = 1 ml $$

Final conditions

$$P_2 = \frac {F}{A}$$ where F is the force you measure to hold the syringe in place at 10 ml and A is the inside cross sectional area of the syringe tube. $$V_2 = 10 ml$$

From these, the values of $c_p$ and $c_v$ for the gas in the syringe, and the appropriate unit conversions for absolute pressure and volume you should be able to calculate the drop in temperature.

Just a last comment about the difference in form of the first law given here and that given by zhutchens1. The former is generally used in physics and thermodynamics while the latter is frequently used in chemistry. In the former work done by the gas is positive whereas in the latter it is negative. As long as we keep that straight there is no difference.

Hope this provides additional help.


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