Work Done by Gas in Carnot Cycle

[Peter G.]

Hi,

I was thinking about the isothermal and adiabatic expansions in the Carnot Cycle and I was wondering:

In these processes, the gas is doing work. In the adiabatic change, the pressure (force), acts over a distance (expansion) transferring energy (Kinetic Energy) - but this is an adiabatic change, so it's impossible to transfer energy. Is it therefore, the process of the pressure expanding the gas that requires the use of energy - in this case, the Kinetic Energy of the molecules?

In the case of the isothermal change (I haven't studied the Second Law of Thermodynamics, but from the looks of it, it seems it isn't possible for this change to happen, but, anyway, ignoring that now...) heat energy is transferred to the gas. This thermal energy would increase the K.E, therefore, increasing the pressure and allowing for the same thing to happen - pressure acting over a distance (doing work) and this action requires energy, energy, in this case, Kinetic, which is equal to the amount of thermal energy transferred in: for ∆Q = 0 + ∆W

In those cases, the energy transfer would be from internal energy/KE to potential energy of the piston?

Are those correct?

 

[ideasrule]

In these processes, the gas is doing work. In the adiabatic change, the pressure (force), acts over a distance (expansion) transferring energy (Kinetic Energy) - but this is an adiabatic change, so it's impossible to transfer energy. Is it therefore, the process of the pressure expanding the gas that requires the use of energy - in this case, the Kinetic Energy of the molecules?

Yes. If the gas expands, its temperature has to go down to compensate.

In the case of the isothermal change (I haven't studied the Second Law of Thermodynamics, but from the looks of it, it seems it isn't possible for this change to happen, but, anyway, ignoring that now...)

There's nothing that prevents isothermal change from happening. If I compress a gas, it would ordinarily heat up. However, if I put it in a refrigerator and cool it at exactly the same rate as it's heating up, I can keep its temperature constant.

heat energy is transferred to the gas. This thermal energy would increase the K.E, therefore, increasing the pressure and allowing for the same thing to happen

No, there can be no change in kinetic energy. Isothermal means constant temperature, and temperature is directly proportional to kinetic energy, so kinetic energy can't change in an isothermal process.

What happens is that pressure and volume may change, but temperature has to stay constant. For example, if I compress some gas, but do it slowly enough that the gas is always in thermal equilibrium with its surroundings, I get an isothermal process. Conversely, if I try to expand some gas, its temperature would ordinarily decrease. However, if heat transfer between the gas and its surroundings is efficient enough, the gas can always stay in thermal equilibrium.

 

[Peter G.]

Ah I think I got it:

1. In the isothermal expansion for example, the gas would still, like in the adiabatic, be expanding at the cost of it's internal energy but, at the same time, immediately, energy is flowing in from the surroundings to compensate?

2. During the compressions, I understand the energy transfer. If we imagine ourselves compressing a gas in a piston, we are exerting a force over a distance and transferring chemical energy in our muscles to internal energy in the gas (or is it potential energy to the piston?)

But how does it happen during expansion? The "act" of the pressure acting on the piston over a distance uses up Kinetic Energy? And then the piston gains, say, potential energy?

Thanks once again,
Peter G.

 

[Andrew Mason]

But how does it happen during expansion? The "act" of the pressure acting on the piston over a distance uses up Kinetic Energy? And then the piston gains, say, potential energy?

During a quasi-static adiabatic expansion, the gas does work on the surroundings in an amount: $W = \int PdV$ where P is the internal pressure of the gas.

If it is not a quasi-static expansion (the external pressure is less than the internal gas pressure by more than an infinitessimal amount), the work done on the surroundings is $W = \int PdV$ where P is the external pressure on the gas (NOT the internal gas pressure).

In both cases, the work is done on the surroundings (e.g. pushing a piston to lift a weight). Since no heat flows into the gas, the energy must come from the internal (kinetic) energy of the gas.

AM

 

[Peter G.]

Hi,

Since no heat flows into the gas, the energy must come from the internal (kinetic) energy of the gas.

Even in an isothermal process?

 

[Andrew Mason]

Even in an isothermal process?

?? An isothermal process is not adiabatic.

In an adiabatic process, dQ = 0. So dU = -dW.

In an isothermal process dQ<>0. So dW <> dU ie. dU = dQ-dW

AM