You are watching: What happens when you compress a gas
$$\fracV_1T_1 = \fracV_2T_2$$
Thus accordingly, during compression, temperature of the gas would certainly decrease. But in Lectures the thedailysplash.tv ,vol 1 by Feynman, that is written:
Suppose the the piston moves inward, so the the atom are gradually compressed into a smaller space. What happens once an atom hits the moving piston? Evidently it picks up speed from the collision. <...> so the atoms are "hotter" as soon as they come out from the piston than they were before they struck it. Because of this all the atoms which room in the vessel will have actually picked up speed. This means that when we compress a gas slowly, the temperature of the gas increases.
(Constant pressure?) So, this is contrary to Charles" law. Why walk this happen? who is right? Or are they both correct? ns am confused. Help.
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thermodynamics pressure temperature ideal-gas volume
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edited Mar 19 "19 in ~ 5:39
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asked Sep 20 "14 in ~ 13:24
$\begingroup$ So over there are various conditions? .... Is no Charles' law speaking that gas'pressure? What is Feynman talk of ? $\endgroup$
Sep 20 "14 in ~ 13:40
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There"s actually no one simple answer to your question, i m sorry is why you are a little bit confused. Come specify your problem fully, you should specify precisely how and also whether the gas swaps heat with its surroundings and also how or also whether the is compressed. Girlfriend should always refer to the full gas law $P\,V=n\,R\,T$ once reasoning. Common situations that are thought about are:
Charles"s Law: The push on the volume gas is constant. No occupational is done by the gas top top its surroundings, nor does the gas do any work ~ above its surroundings or piston or every little thing during any type of change. The gas"s temperature is that of that surroundings. If the approximately temperature rises / falls, heat is transferred right into / out from the gas and also its volume appropriately increases / shrinks so that the gas"s pressure deserve to stay constant: $V = n\,R\,T/P$; v $P$ constant, you deserve to retrieve Charles"s Law;
Isothermal: the gas is compressed / broadened by doing occupational on / permitting its container come do occupational on that is surroundings. You think of it inside a cylinder with a piston. As it go so, warmth leaves / comes right into the gas to store the temperature constant. Together the gas is compressed, the work-related done on it shows up as boosted internal energy, which must be moved to the next site to store the temperature constant. At constant temperature, the gas law becomes $P\propto V^-1$;
Adiabatic: No warm is transferred between the gas and its surroundings together it is compressed / walk work. AGain, you think the the gas in a cylinder v a plunger. This is prototypical case Feynman talks about. As you push on the piston and change the volume $V\mapsto V-\rm dV$, girlfriend do work $-P\,\rm dV$. This energy stays v the gas, so the must present up as boosted internal energy, so the temperature need to rise. Get a bicycle tire pump, organize your finger end the outlet and squeeze the hard and fast through your other hand: you"ll discover you have the right to warm the air up within it quite considerably (put friend lips gently on the cylinder wall surface to feeling the climbing temperature). This situation is defined by $P\,\rm dV = -n\,\tildeR\,\rm d T$. The internal energy is proportional come the temperature and also the variety of gas molecules, and also it is an adverse if the volume rises (in which situation the gas does job-related on the surroundings). But the continuous $\tildeR$ is not the very same as $R$: it relies on the internal degrees of freedom. For instance, diatomic molecules can store vibrational and kinetic power as your bond length oscillates (you deserve to think of them together being organized together through elastic, energy storing springs). So, once we use the gas regulation to remove $P = n\,R\,T/V$ native the equation $P\,\rm dV = -n\,\tildeR\,\rm d T$ we acquire the differential equation:
$$\frac\rm d VV = - \frac\tildeRR\frac\rm d TT$$
which integrates to yield $(\gamma-1)\,\log V = -\log T + \textconst$ or $T\,V^\gamma-1 = \textconst$, where $\gamma=\fracR\tildeR+1$ is referred to as the adiabatic index and is the ratio of the gas"s particular heat at continuous pressure to the particular heat at continuous volume.