Heat, Work, and Energy: Understanding Charles Law and Internal Energy

  • Thread starter Thread starter godingly
  • Start date Start date
  • Tags Tags
    Heat Work
AI Thread Summary
Charles' Law indicates that at constant pressure, increasing the temperature of a gas leads to an increase in volume. The discussion highlights confusion regarding how work and heat affect a gas's temperature, particularly when volume decreases, suggesting that temperature should also decrease. However, it is clarified that the first law of thermodynamics explains that work done on the gas can increase its internal energy, thus raising temperature. Additionally, the atomic-level explanation reveals that kinetic energy transfer occurs through molecular collisions, where fast-moving molecules from a heat source collide with gas molecules, increasing their average kinetic energy. Understanding the assumptions behind gas laws is crucial for applying them correctly in different scenarios.
godingly
Messages
13
Reaction score
0
1) I don't completely understands Charles Law (V = T*k) - This implies that if we increase the volume of an object, it's temperature will rise! Imagine a piston-apparatus - If we push the piston up, and increase the volume of the gas, why would its average kinetic energy rise? What would make the particles (on average) move faster?

2) I've read an example about heat, work and energy. It shows you can change a gas internal energy by either work or heat.
heawor.gif


I don't understand why either of these changes will cause a rise in temperature:
a) In the work case - the gas's volume decreases, so therefore the gas's temperature should decrease as well, not increase! What's happening?
b) In the burner case - I don't understand what happens on the atomic level. I would appreciate an atomic-view explanation of what happens from when you press the burner button to the rise of the gas atoms average kinetic energy.

Thank you,
docendo discimus
 
Science news on Phys.org
Charles' law assumes constant pressure. If you increase the temperature the gas expands under constant pressure. The examples you provided do not have constant pressure so they are not going to follow Charles' law
 
Last edited:
godingly said:
I don't understand why either of these changes will cause a rise in temperature:
a) In the work case - the gas's volume decreases, so therefore the gas's temperature should decrease as well, not increase! What's happening?
The first law of thermodynamics is happening. Q = ΔU - W where W is the work done ON the gas. If Q = 0 (no heat flow into or out of the gas) then ΔU = W. So if there is work done ON the gas, internal energy, U, increases.

b) In the burner case - I don't understand what happens on the atomic level. I would appreciate an atomic-view explanation of what happens from when you press the burner button to the rise of the gas atoms average kinetic energy.
It is all about transfer of kinetic energy by molecular collisions.

The very fast moving molecules in the hot gases in the flame collide with the molecules in the container wall. This gets the molecules in the container moving very fast. The molecules in the container give up some of that kinetic energy to the contents, ie. the gas.

AM
 
Thank you, I understand the formula of the 1st law, but I don't understand :
1) In the case of work, the ideal gas law states PV=nRt. since n & R are constant, and T rises, that means Pressure increases by more than Volume decreases. But Boyle's law states that V and P change inversely proportionally - how is this possible?
2) How, in the microscopic level, would the work increase the average kinetic energy of the atoms? Is it because we increased the pressure on the piston, and now the piston atoms move faster, and then when the gas atoms heat the piston atoms - the gas atoms would move faster?
 
1) Boyle's law assumes the temperature is being kept constant. That's not the case here as you pointed out so Boyle's law doesn't apply. When learning laws such as Crarles' law and Boyle's law you ought to also learn the assumptions that go into the laws so you know when to apply the laws and when not to apply them

2) Yes, that's correct.
 
  • Like
Likes 1 person
godingly said:
2) How, in the microscopic level, would the work increase the average kinetic energy of the atoms? Is it because we increased the pressure on the piston, and now the piston atoms move faster, and then when the gas atoms heat the piston atoms - the gas atoms would move faster?

Just as a moving bat causes a baseball to rebound with more energy than a stationary bat, the moving container wall increases the rebound energy of the molecules that collide with it.

The difference is that with a gas there are many molecules. If the wall is moving slowly, the moving wall increases the speed of each molecule less, but many more molecules are affected because the wall is moving for a longer time.

AM
 
I was watching a Khan Academy video on entropy called: Reconciling thermodynamic and state definitions of entropy. So in the video it says: Let's say I have a container. And in that container, I have gas particles and they're bouncing around like gas particles tend to do, creating some pressure on the container of a certain volume. And let's say I have n particles. Now, each of these particles could be in x different states. Now, if each of them can be in x different states, how many total...
Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'
Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...
Back
Top