Thermodynamics reversible isothermal process

In summary, a system with 10kg of water undergoes a reversible isothermal process with an initial state of saturated vapor at 300°C and a final pressure of 1MPa. The heat transfer during this process is calculated to be 2362.7kJ. To determine the equilibrium vapor pressure of water at 300°C, first assume that liquid water is incompressible. Then, compare the vapor pressure to the system's pressure of 1MPa and calculate the energy needed to change 10kg of saturated vapor to saturated liquid at 300°C.
  • #1
nakas12
12
0

Homework Statement


A system containing 10kg of water undergoes a reversible isothermal process. The initial state can be characterized as saturated vapor at 300°C. The pressure in the final state is 1MPa. The heat transfer during the process in (inKJ)


Homework Equations



Q = m(U2-U1)+W
W = PdV
v1 = V/m
v1 = specific volume at initial state
V = volume

The Attempt at a Solution


The specific volume values were found using the thermodynamic tables for water. I figured that since it was an isothermal process and would be adiabatic. I wanted somebody to see if my answer is correct.

v1=.02167
v2=.25794
V1= 2.5795
V2=2.5794
mass=1.91
T1 =T2 =300°C
Q = mCvo(T2-T1)+W
Q = .717(300-300)+PdV

Q = 0 +1000(2.5794-.2167)
Q = 2362.7
 
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  • #2
What is the equilibrium vapor pressure of water at 300C? How does that compare with a pressure of 1 MPa? What is the volume of the water vapor at the equilibrium vapor pressure?
 
  • #3
I would go as follows:

1. Assume liquid water is incompressible.
2. What is the vapor pressure of saturated vapor/water at 300C? Is it above or below 1 MPa?
3. What's it take in joules to change 10 kg of saturated vapor to saturated liquid (at 300C)?
 

1. What is a reversible isothermal process in thermodynamics?

A reversible isothermal process is a type of thermodynamic process in which the temperature of the system remains constant throughout the process. This means that the heat transfer into or out of the system is balanced by the work done by or on the system, resulting in no change in internal energy.

2. How is a reversible isothermal process different from an irreversible isothermal process?

In a reversible isothermal process, the system is in thermodynamic equilibrium throughout the process, meaning that the process can be reversed without any energy loss. In contrast, an irreversible isothermal process involves a loss of energy due to irreversibilities, such as friction or heat transfer to the surroundings.

3. What is the equation for work done in a reversible isothermal process?

The equation for work done in a reversible isothermal process is W = nRTln(V2/V1), where n is the number of moles of gas, R is the gas constant, T is the temperature in Kelvin, and V1 and V2 are the initial and final volumes of the gas.

4. Can a reversible isothermal process occur in real-life systems?

No, a reversible isothermal process is an idealized concept and cannot occur in real-life systems. This is because it requires infinitely slow processes and perfectly insulated systems, which are not achievable in the real world.

5. How does a reversible isothermal process relate to the second law of thermodynamics?

A reversible isothermal process is a special case of a reversible process, which is a process that does not produce any entropy. This aligns with the second law of thermodynamics, which states that the total entropy of a closed system can never decrease over time.

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