Quasi-Static Processes: Does dQ = 0?

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In summary, in quasi-static processes, the equation dQ = dE + dW still holds, but in this case, dW = -dE. This means that in quasi-static processes, no heat (Q) is absorbed or given off, resulting in dQ = 0. This is because the work done by the system is equal and opposite to the change in internal energy. However, this is not the case for all quasi-static processes, such as an isothermal quasi-static process where dQ is not equal to zero. In Reif's Fundamentals of Statistical and Thermal Physics, it is stated that dW = -dE for quasi-static processes, but this may not necessarily be true for all cases.
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manhattan_project
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We know that dQ = dE + dW for any system. However, in quasi-static processes, dW = -dE. Does this mean that dQ = 0 and no heat (Q) is absorbed or given off? If so, why is that?
 
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manhattan_project said:
We know that dQ = dE + dW for any system.
However, in quasi-static processes, dW = -dE.
Who says that this is the case for a quasi-static process? What about an isothermal quasi-static process, where dQ is not equal to zero?
 
  • #3
I guess I am confused when dQ = 0.
In Reif's Fundamentals of Statistical and Thermal Physics, in Section 2.9 on quasi-static processes, it states that the work dW done by the system when it remains in a particular state r is defined as dWr = -dEr. I understood this as saying that dW = -dE for quasi-static processes
 
  • #4
manhattan_project said:
I guess I am confused when dQ = 0.
In Reif's Fundamentals of Statistical and Thermal Physics, in Section 2.9 on quasi-static processes, it states that the work dW done by the system when it remains in a particular state r is defined as dWr = -dEr. I understood this as saying that dW = -dE for quasi-static processes
I don’t have a copy of that book. But, it dorsn’t sound correct to say that a system remains in a particular thermodynamic state if its internal energy changes. The equation you have written is correct only for an adiabatic change.
 

Related to Quasi-Static Processes: Does dQ = 0?

1. What is a quasi-static process?

A quasi-static process is a thermodynamic process in which the system changes its state very slowly, allowing it to remain in thermal equilibrium with its surroundings at all times.

2. How is a quasi-static process different from other thermodynamic processes?

In a quasi-static process, the system is always in thermal equilibrium with its surroundings. This means that the temperature and other thermodynamic variables such as pressure and volume remain constant throughout the process. In other thermodynamic processes, these variables may change significantly.

3. Does dQ always equal 0 in a quasi-static process?

In a quasi-static process, dQ (the change in heat) is equal to 0 because the system is always in thermal equilibrium with its surroundings. This means that there is no net transfer of heat between the system and its surroundings, and any heat added or removed from the system is done so infinitesimally slowly.

4. Can a quasi-static process be reversed?

Yes, a quasi-static process can be reversed. This is because in a quasi-static process, the system is always in thermal equilibrium with its surroundings, meaning that the process is reversible. This is in contrast to other thermodynamic processes, which may not be reversible.

5. How are quasi-static processes used in thermodynamics?

Quasi-static processes are used in thermodynamics to simplify calculations and analyses of systems. They allow for the application of thermodynamic laws and principles in a more straightforward manner, as the system is always in thermal equilibrium with its surroundings, making it easier to determine the changes in thermodynamic variables such as temperature, pressure, and volume.

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