Thermodynamics- Piston problem, pressure, internal energy

In summary: Delta S=nR\ln 2$$In summary, the conversation discusses various equations and concepts related to a piston in equilibrium and changes in pressure, temperature, and internal energy. The solutions to parts a-c involve using Newton's second law, the ideal gas law, and the equation for change in entropy. The final solution for part c involves using the number of moles of gas, the change in volume, and the gas constant.
  • #1
Ashley1nOnly
132
3

Homework Statement



AshleypastedImage.png


Homework Equations


F= ma
P= F/A

The Attempt at a Solution


a). Assuming the piston is in equilibrium, I'm applying Newton's second law
F=ma=0
Equals zero because it is not moving

Note:
P=F/A which I rearrange to solve for the force for F= PA. This represents the pressure within (pushing upwards).
F=mg which represents the weight of the piston
P=F/A where F=P(atm)A for the outside pressure at room temperature
A=cross-section area of the piston

Now adding up all of the forces
+PA-mg-P(atm)A=0
Now solving for pressure within
P= (mg+P(atm)A)/A = mg/A +P(atm)
Where A is the area of the cylinder so,
/P= (mg)/(pi r^2)

b.) (del)U= (3/2)nR (del)T
There is no change in temperate so,
internal energy =0
therefore
(del)U=0=Q+W

c.)
-Isothermal process, the gas expands and pressure falls while maintaining a constant temperature
-final temperature =initial temperature
-final internal energy- 0 because there was no change in temperature
-change in entropy
(del)S= Q/T
or
(del)S =Q=W
where W= P(del)V[/B]
 

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  • #2
I think they expect you to assume that the pressure outside the cylinder is zero. Under these circumstances, your answer to part (a) would reduce to
$$p_0=\frac{Mg}{A}$$
In part (b) they are asking for the absolute value of the initial internal energy which, according to the standard equation for a monoatomic gas would be:
$$U_0=\frac{3}{2}RT_0$$ So, you were kind of right about this.

Regarding part (c): The weight of the piston is suddenly reduced from Mg to Mg/2. Do you think that the expansion that occurs as a result of this is spontaneous or reversible? Your determination that the final temperature and the internal energy do not change is correct. Your equation for determining the work is OK if P=Mg/(2A). But there is another way of determining the work that the gas does on the piston. From the ideal gas law, In terms of ##V_0##, what is the final volume of the gas? What is the change in volume? What is the displacement of the piston? What is the change in potential energy of the piston if its weight is Mg/2? (This is also equal to the work)

In terms of ##p_0##, ##V_0##, and ##T_0##, what is the number of moles of gas in the cylinder?

Your equation for the determination of the entropy change is incorrect. This equation only applies to a reversible path between the initial and final states of the system. But, the path for this process is not reversible. Can you think of a reversible path between the initial and final states that can be used to determine the entropy change?
 
  • #3
upload_2017-12-15_21-25-52.png


Or

Del(S)=Rln(V(final)/V(initial))
This would work
 

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  • #4
Ashley1nOnly said:
View attachment 216874

Or

Del(S)=Rln(V(final)/V(initial))
This would work
Excellent! Don't forget to multiply by the number of moles.
 
  • #5
Chestermiller said:
Excellent! Don't forget to multiply by the number of moles.
Del(S)=nRln(V(final)/V(initial))

Where
n =moles
V = volume
R= Gas constant
 
  • #6
Ashley1nOnly said:
Del(S)=nRln(V(final)/V(initial))

Where
n =moles
V = volume
R= Gas constant
$$n=\frac{p_0V_0}{RT_0}$$

$$\frac{V_f}{V_i}=2$$
 

Related to Thermodynamics- Piston problem, pressure, internal energy

1. How does a piston work in thermodynamics?

A piston is a component of a thermodynamic system that helps to convert heat energy into mechanical work. It is a cylindrical object that is free to move within a closed container, such as a cylinder. When heat is applied to the system, the pressure inside the container increases, which causes the piston to move. This movement can then be harnessed to perform mechanical work.

2. How does pressure affect thermodynamics?

In thermodynamics, pressure is a crucial factor that affects the behavior of a system. An increase in pressure can result in a decrease in volume, leading to an increase in temperature. This relationship is described by the ideal gas law, which states that pressure and volume are inversely proportional when temperature and the number of particles are held constant.

3. What is internal energy in thermodynamics?

Internal energy is the total energy contained within a system, including its particles and their interactions. In thermodynamics, it is denoted by the symbol U and is a combination of the system's kinetic energy (related to the motion of particles) and potential energy (related to their positions). Internal energy is crucial in understanding heat transfer and work done in a system.

4. How do you solve a piston problem in thermodynamics?

To solve a piston problem in thermodynamics, you need to first identify the variables given in the problem, such as pressure, volume, and temperature. You can then use the ideal gas law or other relevant equations to calculate the unknown variables. It is essential to pay attention to units and ensure they are consistent throughout your calculations.

5. How is thermodynamics related to the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern the behavior of thermodynamic systems. They provide a framework for understanding energy and its transformations in a system. Thermodynamics uses these laws to study and analyze the behavior of various systems, such as engines and refrigerators, and how they convert energy from one form to another.

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