[Physics] Calculating work done on an ideal gas

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I am trying to calculate the work done on an ideal gas in a piston set up where temperature is kept constant. I am given the volume, pressure and temperature.

I know from Boyle's law that volume is inversely proportional to pressure that is,

$$V \propto \frac{1}{p}$$

using this I can calculate the two volumes I need for this equation to calculate work done:

$$\Delta W = – \int^{V_2}_{V_1} p(V)dV$$

but what I do not understand is how to use this equation to help me calculate the work done, I think I am confused by the fact that I need to have $p(V)$ but I am not sure what this is. If you could help me to understand this, that would be great

Best Answer

Try the ideal gas law

$$ p V = N k_B T \Leftrightarrow p = N k_B \frac{T}{V} $$

since $N$, $k_B$ and $T$ are constant, we have

$$ \Delta W = - N k_B T \int_{V_1}^{V_2} \frac{\textrm{d}V}{V} = - N k_B T \left( \ln(V_2) - \ln(V_1)\right) $$

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[Physics] What happens to the temperature when an ideal gas is compressed

There's actually not one simple answer to your question, which is why you are a bit confused. To specify your problem fully, you must specify exactly how and whether the gas swaps heat with its surroundings and how or even whether it is compressed. You should always refer to the full gas law $P\,V=n\,R\,T$ when reasoning. Common situations that are considered are:

Charles's Law: The pressure on the volume gas is constant. No work is done by the gas on its surroundings, nor does the gas do any work on its surroundings or piston or whatever during any change. The gas's temperature is that of its surroundings. If the ambient temperature rises / falls, heat is transferred into / out from the gas and its volume accordingly increases / shrinks so that the gas's pressure can stay constant: $V = n\,R\,T/P$; with $P$ constant, you can retrieve Charles's Law;
Isothermal: the gas is compressed / expanded by doing work on / allowing its container to do work on its surroundings. You think of it inside a cylinder with a piston. As it does so, heat leaves / comes into the gas to keep the temperature constant. As the gas is compressed, the work done on it shows up as increased internal energy, which must be transferred to the surroundings to keep the temperature constant. At constant temperature, the gas law becomes $P\propto V^{-1}$;
Adiabatic: No heat is transferred between the gas and its surroundings as it is compressed / does work. AGain, you think of the gas in a cylinder with a plunger. This is prototypical situation Feynman talks about. As you push on the piston and change the volume $V\mapsto V-{\rm d}V$, you do work $-P\,{\rm d}V$. This energy stays with the gas, so it must show up as increased internal energy, so the temperature must rise. Get a bicycle tyre pump, hold your finger over the outlet and squeeze it hard and fast with your other hand: you'll find you can warm the air up inside it quite considerably (put you lips gently on the cylinder wall to sense the rising temperature). This situation is described by $P\,{\rm d}V = -n\,\tilde{R}\,{\rm d} T$. The internal energy is proportional to the temperature and the number of gas molecules, and it is negative if the volume increases (in which case the gas does work on its surroundings). But the constant $\tilde{R}$ is not the same as $R$: it depends on the internal degrees of freedom. For instance, diatomic molecules can store vibrational as well as kinetic energy as their bond length oscillates (you can think of them as being held together by elastic, energy storing springs). So, when we use the gas law to eliminate $P = n\,R\,T/V$ from the equation $P\,{\rm d}V = -n\,\tilde{R}\,{\rm d} T$ we get the differential equation:

$$\frac{{\rm d} V}{V} = - \frac{\tilde{R}}{R}\frac{{\rm d} T}{T}$$

which integrates to yield $(\gamma-1)\,\log V = -\log T + \text{const}$ or $T\,V^{\gamma-1} = \text{const}$, where $\gamma=\frac{R}{\tilde{R}}+1$ is called the adiabatic index and is the ratio of the gas's specific heat at constant pressure to the specific heat at constant volume.

[Physics] Confusion in the pressure-volume work in an irreversible process

For $PV$ work in which $P_{\text{ext}}$ is a constant, you could actually intepret it as $W=Fx$ where $F$ is the force on the piston (due to pressure) and $x$ is the distance travelled by the piston.

From here you can see that the force which you applied plays no part in determining the work done against opposing force.

Imagine lifting a mass off the ground, it does not matter how much force you exert in lifting the mass, all that matters is the distance the mass is lifted, as well as the gravitational force, mg. The work done is simply $W=mgh$. More explanation could be found here

Similarly, in this case, the internal pressure plays no part in determining the work done against an external pressure, as the work done against a constant opposing force(or opposing pressure) is only determined by the distance travelled by the piston (or change in volume).

Do note that this is true if and only if the piston is stationary before and after expansion. If expansion causes the piston to be moving in the final state, work done may include KE term.

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