# Adiabatic processes and the Ideal Gas Law [closed]

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 **

## closed as off-topic by Mithoron, andselisk♦, Tyberius, tschoppi, WildcatJan 26 at 20:41

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• I think you should consider the consequences of the word "adiabatic" in more detail and come back to edit your question with that information. – tschoppi Jan 25 at 9:52

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 $$\gamma > 1 .(\gamma = \frac{C_p}{C_V})$$ 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.

In an adiabatic expansion, if the volume increases, the pressure decreases, but the pressure decrease (ratio) is more than the volume increase (ratio), so the temperature also decreases to make good on the ideal gas law. If the volume decreases, the pressure increases, but the pressure increase(ratio) is more than the volume decrease (ratio), so the temperature also increases, to make good on the ideal gas law.

Under adiabatic conditions, heat evolved is zero. Considering first law of thermodynamics, any work performed by the gas or on it would involve expenditure form its internal energy.

If the gas expands, it is doing negative work, hence change of internal energy is negative. So $$T(initial)> T(final)$$.

Ideal gas law is infact a summation of Boyle's, Charle's and Avogrado's laws which describes the state in which the gas exist. Under adiabatic conditions, pressure changes along with volume and temperature. So with increasing volume, $$T/P$$ also increases. The extent of decreasing $$P$$ is greater than decreasing $$T$$, due to the factor $$γ$$.

$$P^{1-\gamma}T^\gamma=\text{constant}$$

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.

If you have a box divided into two parts, and one part is filled with an ideal gas, the particles will have some average kinetic energy. If you now remove the divider, the particles have more room available to them, but they will not speed up. Instead, the pressure will decrease because less particles will hit a unit area of the wall in some time unit (because there is more space, it will take longer for an individual particle to make a round trip hitting the wall once and then hitting it again).

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.

When a particle hits a stationary wall, it bounces off without a change in kinetic energy. If it bounces off a wall that can move, it slows down because some momentum is transferred to move the wall. Try throwing a tennis ball against a concrete wall vs against the side of a cardboard box. (Or take a look how one bowling ball is stopped by another.)

the ideal gas law which says that the volume of the gas is proportional to the temperature and vice versa

The ideal gas law says that V and T are proportional if everything else (n and P) is constant. However, by removing some pebbles from the piston, you change the pressure the piston exerts on the gas, and once the system is back in equilibrium, the pressure inside the canister will also be lower. So V increases as P decreases. In an isothermal process, T would stay constant (particles regain lost kinetic energy from the surrounding) and with it PV. In an adiabatic process, T decreases as well, and you have to look at the specific parameters to calculate how much they change.