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# Work done by isothermic process

Isothermic and adiabatic processes. Calculating the work done by an isothermic process and seeing that it is the same as the heat added. Created by Sal Khan.

## Want to join the conversation?

• I'm still a bit confused why the fact that temperature is kept constant implies that the change in internal energy is 0?
• Temperature is directly proportional to average kinetic energy. Internal energy equals (3/2)nRT, so the only difference between temperature and internal energy is a scaling factor of (3/2)nR.
If temperature of a system is kept constant, that means that the kinetic energy of the particles in that system is also constant. In an ideal gas, internal energy depends solely on kinetic energy, so if temperature changes, internal energy also changes proportionally.
• Its still possible to solve for this though without the fact of using Calculus right?

in other words, you can still solve this problem using simple algebra.
• Not that I'm aware of. Somewhere in there you will have to take a limit. When you take the limit of something, you enter the world of calculus. There is no way around it that I'm aware of. The Internet might surprise us both all the same.
• Is there any reason as to why P-V curve is a hyperbola ?? or just an observation ??
• the reason why p-v curve is a hyperbola is because p is inversely proportional to v. And as Sal says in the video the graph of y=1/x is a rectangular hyperbola. That is the same as p=nrt where n,r,t are constants so the graph of p-v comes out to be a rectangular hyperbola as well. See?
• Explain the difference between Temprature and Heat?
• Heat is energy, whereas temperature is a measure of energy. (From coolcosmos.edu) "Heat is the total energy of molecular motion in a substance while temperature is a measure of the average energy of molecular motion in a substance. Heat energy depends on the speed of the particles, the number of particles (the size or mass), and the type of particles in an object. Temperature does not depend on the size or type of object." Hope that helps!
• What happens at the microscopic level when the kinetic energy of the ideal gas is transferred to the piston?

We're saying that the total energy in an ideal gas (the internal energy) is the average kinetic energy of its elementary entities (atoms), and that they only interact through elastic collisions with no loss of energy. So if this is true, then how do the atoms of the ideal gas transfer energy to a piston if when colliding with the piston and moving it they themselves do not lose energy? Or are we saying that specifically for situations where on the macroscopic scale we know there is a transfer in energy, then the microscopic collisions of the atoms become non-elastic and thus transfer energy?
• They do lose energy, if the piston is allowed to move. That's how the piston moves. That doesn't mean the collision is not elastic. Elastic collisions conserve KE but they allow KE to be transferred from one body to another.
• how is the slope under the ov graph equal to work??
thanxx
• Work is force times distance. Pressure is force / area. Multiply pressure times volume and you will get Newton*meters.
• Intuitively I'm having trouble with the temperature having to go down in an adiabatic system. While I understand the math requires it, it seems to me that by simply moving one of the walls that I'm NOT changing the kinetic energy of any of the particles in the system. That they all have the same velocities as before and therefore the overall kinetic energy of the closed system is the same. Is it because the actual movement of the wall is caused by some transfer of momentum from the particles to the moving wall therefore causing a reduction in kinetic energy?
• You are changing the KE of the particles.
Think of a tennis ball hitting a racket. Have you ever seen someone "catch" a ball by letting the ball hit the racket while moving the racket backwards? That's what happens when a molecule hits a wall that then moves backwards - it bounces off with less speed than it had before it hits. momentum and KE is transferred to the wall FROM the molecule.

It's just the exact opposite of what happens when the wall (or racket) is moving toward the molecule (or ball) and the molecule (ball) leaves faster than it came in. In that case momentum and KE is transferred TO the molecule (ball)

There is a good simulation on www.phet.colorado.edu where you can play with volume and temperature and pressure and you can even watch the individual molecules moving around.
• At Sal asked a follow up question and later explained that some amount of heat must be added to the system which would be equal to the amount of work done. I have 2 questions

1) The work is done by or on the system?
2) In my textbook it states that work done would be equal to negative heat, and negative work done would be equal to positive heat. Why so? In the video Sal says it is just equal to the work done?

Sorry if the question seems silly.