Delta U, H, S for an Irreversible Adiabatic process.

In summary, for the irreversible expansion against a constant pressure of 0.5 atm, the values of ΔU, ΔH, ΔSsystem, ΔSsurroundings, and ΔStotal can be calculated using the first law of thermodynamics and the equations for work and entropy change.
  • #1
cromat
1
0
Suppose that one mole of a monatomic perfect gas at 27°C and 1.00 atm pressure
is expanded adiabatically (i.e. no heat transfer, so that the temperature must fall) in two
different ways: (a) reversibly, to a final pressure of 0.5 atm, and (b) against a constant
external pressure of 0.5 atm.
(1) One point each. For the reversible case, determine the values of ΔU, ΔH, ΔS system,
ΔS surroundings, and ΔS total.
(2) Do the same for the irreversible expansion against a constant pressure of 0.5 atm.

I think I'm fine with part A, but I am completely lost for part B and can't really start because I have no idea what formulas to use or where to even start.

Unlike part A where ΔU=pdV=CdT and so allow you to integrate with respect to both sides, in part B the pressure is constant and so I'm not sure if I'm supposed to integrate the above formula.

I first thought that the change in volume in the first part would be the same in the second part, which would allow me to easily find ΔU, but it seems this is not the case.

I'm not looking for someone to do my work for me, but just lead me to some formulas concerning all the Δ's and show me where to start.

Thanks.
 
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  • #2
For part B, you can use the first law of thermodynamics, which states that ΔU = q + w. Since there is no heat transfer (adiabatic process), q = 0. The work done by the system is equal to the negative of the change in the internal energy, so w = -ΔU. Therefore, ΔU = -w. The work done against a constant external pressure is given by w = -Pext*ΔV, where Pext is the external pressure and ΔV is the change in volume. Therefore, ΔU = Pext*ΔV. You can then calculate the other parameters by using the equations for ΔH, ΔSsystem, ΔSsurroundings, and ΔStotal.
 
  • #3


I am happy to help you with this problem. Let's start by breaking down the problem and understanding the concepts involved.

First, let's define the terms ΔU, ΔH, and ΔS. ΔU is the change in internal energy of a system, ΔH is the change in enthalpy, and ΔS is the change in entropy. In this problem, we are dealing with an irreversible adiabatic process, meaning there is no heat transfer and the system is insulated. This also means that the change in enthalpy, ΔH, is equal to the change in internal energy, ΔU, since there is no work being done on or by the system.

Now, let's consider the two different ways the gas is expanded: (a) reversibly and (b) against a constant external pressure. In the reversible case, the expansion is done slowly and the system is in equilibrium at all times, meaning the process is reversible. In the irreversible case, the expansion is done quickly and the system is not in equilibrium, making the process irreversible.

For (a) the reversible case, we can use the ideal gas law, PV = nRT, to determine the change in volume, ΔV, of the gas. Since the temperature is constant at 27°C, the change in enthalpy, ΔH, is equal to the change in internal energy, ΔU. We can then use the first law of thermodynamics, ΔU = Q - W, to calculate the change in internal energy. Since the process is adiabatic, there is no heat transfer, so Q = 0. And since the process is reversible, the work done is equal to the area under the curve on a PV diagram. This can be calculated using the formula W = -PΔV. Finally, the change in entropy, ΔS, can be calculated using the formula ΔS = Q/T. Since Q = 0, ΔS = 0 for the system. However, for the surroundings, there is a decrease in entropy due to the work done on the surroundings, so ΔS surroundings < 0. And for the total system, ΔS total = ΔS system + ΔS surroundings = 0 + (-ΔS surroundings) = ΔS surroundings. So, in summary:

ΔU = 0
ΔH = 0
ΔS system = 0
 

Related to Delta U, H, S for an Irreversible Adiabatic process.

1. What does "Delta U" stand for in an irreversible adiabatic process?

"Delta U" represents the change in internal energy of a system during the process. It is a measure of the energy transferred to or from the system, and is often expressed in joules (J) or kilojoules (kJ).

2. How is "Delta H" different from "Delta U" in an irreversible adiabatic process?

"Delta H" represents the change in enthalpy of a system during the process. It takes into account not only the internal energy, but also the work done by or on the system. In an adiabatic process, where no heat is exchanged, "Delta H" is equal to "Delta U".

3. What does "Delta S" indicate in an irreversible adiabatic process?

"Delta S" represents the change in entropy of a system during the process. It is a measure of the disorder or randomness of the system. In an irreversible adiabatic process, the entropy of the system generally increases.

4. How are "Delta U", "Delta H", and "Delta S" related in an irreversible adiabatic process?

In an irreversible adiabatic process, where no heat is exchanged with the surroundings, "Delta U" is equal to "Delta H", and "Delta S" is greater than 0. This means that the internal energy and enthalpy of the system remain constant, while the entropy increases.

5. Can "Delta U", "Delta H", and "Delta S" ever be negative in an irreversible adiabatic process?

Yes, it is possible for "Delta U", "Delta H", and "Delta S" to be negative in an irreversible adiabatic process. This would indicate a decrease in internal energy, enthalpy, and entropy of the system, respectively. However, in an adiabatic process, it is more common for these values to be positive or remain constant.

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