T vs V for adiabatic processes

  • Thread starter RadiumBlue
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In summary: No, it's not. So that can't be right.Instead, use the fact that the process is adiabatic, which means that E is constant with respect to changes in T and V. So:dE = 0Use your equation for E to express that in terms of dT and dV.Then, use the equation for p:p = (1/3)σT^4to express dV in terms of dT and dE, and solve for the desired relationship.
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RadiumBlue
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Homework Statement


[/B]
a) Black body radiation:

The energy and pressure of black body radiation depend on [itex]T[/itex] and [itex]V[/itex] as [itex] E = σV T^4 [/itex] , [itex]p=\frac{1}{3}σT^4 [/itex] σ = a constant

(1) Suppose that the temperature and volume of a box of radiation change adiabatically, which means that there is no heat flow. First, find the relation between [itex]dE[/itex] and [itex]dT[/itex] in this process. Next using equation (1) show that [itex]T ∝ V^-\frac{1}{3}[/itex]

b)Pressureless glop

Suppose that the energy and pressure of a different substance are given by [itex] E = γV T^\frac{2}{3}[/itex] ,[itex] p = 0 [/itex] γ = constant

The temperature and volume of a box of this substance are changed adiabatically. What is the relation between T and V during this process?

Homework Equations


See problem description

The Attempt at a Solution


I'm not quite sure where to begin with this one. I've solved every other problem in the problem set without a problem, but this one I'm a little confused where to start.

[itex] \Delta E = Q - W [/itex]

The processes are adiabatic, so Q = 0. Also, [itex]W = pdV[/itex]

Therefore
[itex] \Delta E = -pdV [/itex]

Is it asking me to find dE/dT? For part a this would be:
[itex]\frac{dE}{dT} = 4σVT^3 [/itex]

After I get these pieces, I'm not sure how to assemble them into the answers the problems are asking for.
 
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  • #2
RadiumBlue said:
##\frac{dE}{dT} = 4σVT^3##
That would only be true if V is constant with respect to changes in T. Is it?
 

Related to T vs V for adiabatic processes

1. What is the relationship between T and V in an adiabatic process?

In an adiabatic process, the temperature and volume of a system are inversely related. This means that as the volume of a system decreases, the temperature increases, and vice versa.

2. How is the adiabatic process different from the isothermal process?

The main difference between an adiabatic process and an isothermal process is that in an adiabatic process, there is no exchange of heat between the system and its surroundings, while in an isothermal process, the temperature of the system remains constant.

3. Can the adiabatic process be reversible?

Yes, the adiabatic process can be reversible. This means that the system can be returned to its original state by reversing the process without any energy loss. However, in reality, most adiabatic processes are irreversible due to factors such as friction and heat dissipation.

4. How is the adiabatic process related to the first law of thermodynamics?

The adiabatic process is related to the first law of thermodynamics, which states that energy cannot be created or destroyed, but it can be transferred or converted from one form to another. In an adiabatic process, there is no exchange of heat, so the change in internal energy of the system is equal to the work done on the system.

5. What are some practical applications of the adiabatic process?

The adiabatic process has many practical applications, such as in car engines, where the compression of air and fuel causes an increase in temperature and pressure, resulting in the expansion of gases and the movement of the pistons. It is also used in refrigerators and air conditioners, where the compression and expansion of gases are used to cool the surrounding air. In addition, adiabatic processes are used in the manufacturing of certain materials and in industrial processes such as welding and metal forging.

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