www.introphysics.info/IntroToEqThermo/Thermo.pdf
read that before reading anything else on earth.
Physics and Chemistry have opposite conventions over choosing the sign of work done. That is a FACT! (It's not that you are confused)
In physics, work done by the system is taken as positive.
In chemistry, work done on the system is taken as positive.
The only way to escape this stupid convention tricks, is to really know what's happening.
U, internal energy is the sum of potential, kinetic energies, etc of molecules in a gas. (And here's the killer tip. For ideal gases, we neglect the potential energy, etc. So internal energy is directly the kinetic energy. And temperature, as we'll learn in kinetic theory, is a function of kinetic energy. So, internal energy is directly measured by temperature for IDEAL gases)
Q is heat applied. Thankfully it's taken positive when heat is applied to the system in both physics and chemistry.
(And when heat is applied to the system, we are supplying energy to the system, that is to say the internal energy rises)
W is work done. (Let's forget if it's by the system or on the system)
When we try to contract the gas, we hit the molecules and inadvertently add to their kinetic energy. So, internal energy increases when the gas is contracting.
When the gas is expanding, the molecules give up a part of their kinetic energy to push the wall around. So, internal energy decreases on expansion.
When we are pushing a piston as in the textbook, W is PΔV only. You don't need to add VΔP. We are not differentiating. We are only finding FΔs
F=PA (actually the P is the external pressure. Can't type subscripts, that's why)
AΔs=V
So, W = P A Δs = P ΔV
ΔU=ΔQ+ΔW as per chemistry textbook.
ΔU=ΔQ-ΔW as per physics textbook.
Both read the same, as the sign of ΔW would be opposite in both.
Now, let's see what happens during thermodynamic processes.
Isothermal expansion or compression.
When compressed, as I told earlier, the internal energy should increase.
But since it's isothermal the internal energy can't increase. (ΔU=0)
So, what do we do? Give out that energy as heat!
And during expansion that much heat energy is absorbed as much as is expended in expanding.
Let's try to quantify this heat released (or the work done) in terms of P, V, etc.
Since it's isothermal, T is a constant. And we know that PV=nRT
Now when we are compressing the gas, we are decreasing the volume from V1 to V2. That SHOULD be accompanied by a proportional increase in pressure from that formula.
Now we've found W=PΔV. Can we just plug in P as the pressure right now, and ΔV as V2-V1 ??? (The P would be external atmospheric pressure)
Well, that's what happens when a gas bomb explodes. All it takes is an instant, and it'd expand to its final size.
Extra reading
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/CarnotEngine.htm
read that before reading anything else on earth.
Physics and Chemistry have opposite conventions over choosing the sign of work done. That is a FACT! (It's not that you are confused)
In physics, work done by the system is taken as positive.
In chemistry, work done on the system is taken as positive.
The only way to escape this stupid convention tricks, is to really know what's happening.
U, internal energy is the sum of potential, kinetic energies, etc of molecules in a gas. (And here's the killer tip. For ideal gases, we neglect the potential energy, etc. So internal energy is directly the kinetic energy. And temperature, as we'll learn in kinetic theory, is a function of kinetic energy. So, internal energy is directly measured by temperature for IDEAL gases)
Q is heat applied. Thankfully it's taken positive when heat is applied to the system in both physics and chemistry.
(And when heat is applied to the system, we are supplying energy to the system, that is to say the internal energy rises)
W is work done. (Let's forget if it's by the system or on the system)
When we try to contract the gas, we hit the molecules and inadvertently add to their kinetic energy. So, internal energy increases when the gas is contracting.
When the gas is expanding, the molecules give up a part of their kinetic energy to push the wall around. So, internal energy decreases on expansion.
When we are pushing a piston as in the textbook, W is PΔV only. You don't need to add VΔP. We are not differentiating. We are only finding FΔs
F=PA (actually the P is the external pressure. Can't type subscripts, that's why)
AΔs=V
So, W = P A Δs = P ΔV
ΔU=ΔQ+ΔW as per chemistry textbook.
ΔU=ΔQ-ΔW as per physics textbook.
Both read the same, as the sign of ΔW would be opposite in both.
Now, let's see what happens during thermodynamic processes.
Isothermal expansion or compression.
When compressed, as I told earlier, the internal energy should increase.
But since it's isothermal the internal energy can't increase. (ΔU=0)
So, what do we do? Give out that energy as heat!
And during expansion that much heat energy is absorbed as much as is expended in expanding.
Let's try to quantify this heat released (or the work done) in terms of P, V, etc.
Since it's isothermal, T is a constant. And we know that PV=nRT
Now when we are compressing the gas, we are decreasing the volume from V1 to V2. That SHOULD be accompanied by a proportional increase in pressure from that formula.
Now we've found W=PΔV. Can we just plug in P as the pressure right now, and ΔV as V2-V1 ??? (The P would be external atmospheric pressure)
Well, that's what happens when a gas bomb explodes. All it takes is an instant, and it'd expand to its final size.
Extra reading
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/CarnotEngine.htm
No comments:
Post a Comment