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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 learned in my Physics class that in an isothermic process the change in internal energy is 0; because deltaU=(3/2)nR(delta T) meaning that Q=(negative work). An adibiatic process is different from that because it implies that Q is 0, so delta U=Work. Why does Sal treat these two different processes as if they were the same?
• This is actually incorrect. In an adiabatic process (as he states)is isolated from the world. No heat goes into or out the system. Thus, Qin=0 and Qout=0. Delta Q does equal zero but also Q=0. For an ideal gas, internal energy is only dependent on temperature, and if the temperature does not change, then the change in internal energy is zero. For the temperature to stay the same we must have some Qin to counterbalance the decrease in temperature from removing the "pebbles." Thus change in temp, delta T=0, thus delta U must be 0. Ex. delta U=(3/2)nR(delta T), delta T is 0, so the right side is equal to zero because you are multiplying the (3/2)nR by 0. This gives you the negative work (or Wout).

So, for adiabatic delt U=work. If you remove pebbles and it expands, the temperature will go down because no heat is added. This is work out, or negative work. If temperature goes down, then internal energy (kinetic energy of particles) also goes down which is shown from the negative work, work out.

It is different because the temperature of the adiabatic system does change because the temp must go down in expansion because no heat(Q) is added. Or, Q=0 so delta T is negative. However, in isothermal, the temperature does not change because heat is added from the surroundings to counteract the temperature drop from the expansion. Or, delta T=0 and Q is positive.

I was in a rush, hope that made sense.
• Around , Mr. Khan explains that the temp. decreases. I am confused why this is the case. I thought that since there is more room in the system to move, velocity would increase? Since temp is the average K.E(=1/2mv^2), wouldn't the temp also increase?
• Temperature decreases because kinetic energy has been converted into work, as the gas does work to expand its volume. And since temperature is proportional to the average Kinetic energy, the temperature has decreased in the adiabatic system. An increase in volume is not related to an increase in velocity of particles
• 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.
• AT U say that if temp wont change the no. of particles r not gonna change???what if the temp. was changing?from where would u change the no.of particles even when delta t is a higher value.i thought that the system was closed....am...well of course it is....i just don't understand that.could u please tell what did u meant 2 say by that..last mathy part was ok.well i know its very silly question but i look forward for any ans.if it may help
• You are right, neither the temperature (T) nor the number of particles (n) change. The number of particle will not change because no particles can come in or go out of the system. The Temperature will not change because the reservoir keeps it from changing.

The equation he gives is only for when the temperature stays the same. If the temperature was changing you could not use this equation.

Hope that helps
• 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.