Energy Change in Reversible Adiabatic Expansion of a van der Waals gas

In summary, the equation of state of a gas is given by Beta p = rho/(1-b rho) - beta a rho^2, where beta=1/kb T, b is a constant, and rho is the molecular density. The internal energy of the gas is U=(5/2)N kb T - N a rho. To determine the final internal energy of the gas, initially at temperature T0, when expanded from volume V0 to 1.1V0, the following three cases must be considered: A. adiabatically with zero external pressure, B. adiabatic while maximizing work done by the system, and C. isothermally. In case A, there is no change
  • #1
whizsix
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0

Homework Statement


Suppose the equation of state of a gas is
Beta p=rho/(1-b rho) - beta a rho^2
where beta=1/kb T, b is a constant and rho is the molecular density, N/V. The internal energy of this gas is given by
U=(5/2)N kb T - N a rho

Determine the final internal energy of the gas, initially at temperature T0, if it is expanded from a volume V0 to 1.1V0. Answer the question for the following three cases expressing your answer in terms of T0,V0, and N

A. adiabatically with zero external pressure
B. adiabatic while maximizing work done by the system
C. isothermally

Homework Equations



lots, mainly dU=dq+dw
dw=-pdV

The Attempt at a Solution


A. is fine: no change in internal energy (free adiabatic expansion)
B. I can't seem to get this. Maximizing work means making the process reversible, I believe. Lots of stuff talks about solving this this for an ideal gas, but trying in the same manner (dU=-pdV, substituting dU/dt dt=Cv dt for dU and an expression for p then integrating) does not seem possible because with the vdW equations, I can't seem to separate T from V to integrate. If someone has some insight, I would greatly appreciate it.
C. I haven't gotten here yet from working on B. but I'm fairly certain this should be fairly simple.
 
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  • #2
I would assume dU=dq, which would be 0 because it is isothermal. Then just pdV=0, so no change in internal energy either.
 

1. What is reversible adiabatic expansion?

Reversible adiabatic expansion is a process in which a gas expands without any heat transfer occurring between the gas and its surroundings. This means that the temperature of the gas remains constant throughout the expansion.

2. What is a van der Waals gas?

A van der Waals gas is a real gas that takes into account the non-ideal behavior of gas molecules, such as their finite size and intermolecular forces, in contrast to the ideal gas law which assumes no intermolecular interactions.

3. How does energy change during reversible adiabatic expansion of a van der Waals gas?

During reversible adiabatic expansion, the energy of the van der Waals gas decreases as the gas expands. This is because the gas is doing work on its surroundings, using up some of its energy in the process.

4. Is the energy change in reversible adiabatic expansion of a van der Waals gas different from that of an ideal gas?

Yes, the energy change in reversible adiabatic expansion of a van der Waals gas is different from that of an ideal gas. This is because the van der Waals gas takes into account intermolecular interactions, which affect the energy of the gas.

5. What is the relationship between pressure and volume in reversible adiabatic expansion of a van der Waals gas?

In reversible adiabatic expansion of a van der Waals gas, the pressure and volume are inversely proportional to each other. This means that as the volume of the gas increases, the pressure decreases, and vice versa.

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