@user8718165 what process are you considering? For example are you applying a force to the piston to hold it in place, then gradually decreasing that force to allow the piston to slide out slowly?
@user8718165 suppose the piston area is $A$, and the pressure inside is $V$, then the force on the piston is just $F=PA$, and the acceleration of the piston is $a = PA/m$, where $m$ is the mass of the piston.
Assuming this is an adiabatic expansion then the pressure and volume are related by $PV^\gamma = C$ for some constant $C$, so $P=C/V^\gamma$. Then our equation becomes:
You could solve that equation to get $x$ as a function of time, then use this to calculate the pressure as a function of time. But I doubt the equation has a simple analytic solution.
@JohnRennie now suppose I try to lift the piston upwards so that at each moment accln of the piston will be higher than what it was in our case when only the gas was pushing on the piston...I won't pull it fast enough so as to create joule expansion...
@user8718165 strictly speaking the pressure is only defined if the expansion is reversible, and an expansion is only reversible if it is infinitely slow.
The reason there is a pressure on the piston is that gas molecules hit the piston and bounce off it. That means the momentum of the gas molecule changes (on average) by $2mv_{rms}$ and that momentum is transferred to the piston.
If the piston is stationary then the collision is elastic i.e. the gas molecule bounces off the piston with the same speed it hit the piston.
But if the piston moves then some of the kinetic energy of the gas molecule is transferred to the piston and the gas molecule bounce off with a lower velocity than it hit. And if the gas molecule velocity falls that means the temperature falls.
So in the instant after a collision the molecules that have just bounced off the piston have a lower velocity than the molecules far away from the piston i.e. our gas no longer has a uniform temperature.
And if the temperature isn't uniform the pressure isn't uniform either.
In practice gas molecules move very fast, so the slower molecules near the piston quickly collide with the hotter molecules far from the piston and the velocities even out again.
So if the piston is moving slowly compared to the speed of the gas molecules then we can treat the pressure as uniform to a very good approximation.
@JohnRennie sir does it mean that if a gas molecule bounces against a very fast moving piston...the molecule will have a very low KE in opposite direction...right
@JohnRennie that means I can change temperature of a volume of gas by moving the piston very fast...but that concept isn't much of a use in thermodynamics...right sir?
In real life gas molecules move so fast that the non-uniformity is too small to matter. However you are right that strictly speaking equilibrium thermodynamics is an idealised case that is only an approximation to the real world.
There is a discipline called non-equilibrium thermodynamics that deals with situations where the non-uniformity is big enough to matter.
Non-equilibrium thermodynamics is a branch of thermodynamics that deals with physical systems that are not in thermodynamic equilibrium but can be described in terms of variables (non-equilibrium state variables) that represent an extrapolation of the variables used to specify the system in thermodynamic equilibrium. Non-equilibrium thermodynamics is concerned with transport processes and with the rates of chemical reactions. It relies on what may be thought of as more or less nearness to thermodynamic equilibrium. Non-equilibrium thermodynamics is a work in progress, not an established edifice...
@JohnRennie In most of the cases I've encountered during expansion...only the gas applies the pressure to the piston and gets colder...no extra force is used to pull the piston....
I was just imagining varois cases....so I asked you
If you go to the bottom of page 3, it gives a diagram of the circuit of interest. The input voltage is the function generator, but I'm a bit confused what the output voltage is. Is this just the voltage of the capacitor?
I always have difficulty remembering if the capacitor voltage leads or lags. Likewise with inductors, though I remember that inductors and capacitors are opposites i.e. one leads and the other lags.
OK, so in that case, one of the questions states: "Does the voltage across the resistor and capacitor add to the voltage from the function generator? Is this what your expected? If they do not add, what is the correct relationship between these terms?" - I'd expect not, because now we have them just adding as vectors.
The other question would be: how would I measure the phase difference from the oscilloscope?
I know there's no button that just lets you measure the phase difference, so I'd have to make some measurements of the time or something but I'm not sure exactly what of.
@JohnRennie the most important part I'm trying to figure out is Q11 - am trying to sketch a rough idea of what I'd expect the phasor diagram to look like
To measure the phase lag you have to make measurements off the oscilloscope display. I would look at the points where the voltage is zero and measure the distance between those points to get the phase lag. Scopes normally have a grid on the display for you to make the measurement.
the last thing I need help with is the phasor diagram
we already determined that the phasor for the output voltage is just the voltage across the capacitor, so that's done. is the phasor for the input voltage "in sync" with the phasor for the voltage across the resistor?
no it can't be
the current should be in sync with the voltage across the resistor though right?
When you add the resistor and capacitor together using your phasor diagram you get the impedance for the pair of them, and the phase angle for the pair is the phase difference between the voltage and current flowing through the battery.
OK, look at that phasor diagram and imagine it was the complex plane i.e. the horizontal axis is the real axis and the vertical axis is the imaginary axis. Then the resistance would be a real number i.e. just $R$.
The inductance would be an imaginary number $iL$ and the capacitance would be an imaginary number $-iC$. OK so far?
So the resistance, inductance and capacitance are adding just like complex numbers. Obvious they aren't really complex numbers but they add as if they were. The complex number is just a way of encoding the resistance and the phase into a single quantity.
Anyhow we can write complex impedances as: $Z_R = R$ $Z_L = i \omega L$ $Z_C = 1/(i \omega C)$
And these complex impedances add just like regular resistances do. e.g. for an inductor, capacitor and resistor in series you just have $Z = Z_R + Z_L + Z_C$
Suppose you have a charge density $\rho(\mathbf r)$. Note that we need a vector to specify the position because the charge may vary in a random way so it isn't enough to just make it a function of distance.
And consider a small volume element $dV$. We could use a cube of sides $dx$, $dy$ and $dx$ in which case $dV = dxdydz$, or some other shape of volume element. It doesn't matter.
Aladdin's question was about an electron, so the charge is $-e$. If we consider the point $A$, with distance to the charges $r_1$ and $r_2$, then we just add the potentials for the two charges. So we get:
$$ V = \frac{k(-e)(-q)}{r_1} + \frac{k (-e) (+q)}{r_2} $$
Well the argument is always negative so there are no real values of the function. The function can be defined if you allow complex numbers, though taking logs of complex numbers gets a little complicated.