Solve Thermodynamics Problem: Expansion Irreversibly and Isothermally at 27°C

In summary, the entropy change of the system during the isothermal expansion is 13.4 J/K/mol, while the entropy change of the surroundings is -13.408 J/K/mol. This difference is due to the work done by the system on the surroundings during the expansion process. When determining entropy changes, it is important to consider both the energy absorbed and the work done by the system.
  • #1
sachin123
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0
Expansion irreversibly and isothermally of an ideal has from a volume of 2dm3 to 10 dm3 against a const pressure of 1 atm at 27 C .Find entropy change of the system and also surroundings due to the process.

Heres my attempt:
delta S=energy absorbed/temp
for the system,
= energy absorbed/temp=RTln(V2/V1)/T(as it is an isothermal exp)
putting in values I got,13.4J/K/mol
since energy absorbed by surroundings= -(energy absorbed by system)
delta S of surroundings= -13.4J/K/mol
My book gives different answer fro 2nd part.
Why am I wrong?
I just don't understand when we have to consider work and when not?(like for the system or the surroundings?)
 
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  • #2


Your attempt at finding the entropy change of the system is correct. However, the entropy change of the surroundings should also take into account the work done by the system on the surroundings. In this case, the system is doing work on the surroundings as it expands against a constant pressure. The work done by the system is given by W = PΔV, where P is the pressure and ΔV is the change in volume. In this case, ΔV = 8 dm3 = 8 L = 0.008 m3. Therefore, W = (1 atm)(0.008 m3) = 0.008 J.

The total entropy change of the surroundings is the sum of the energy absorbed by the surroundings and the work done by the system on the surroundings. Therefore, the entropy change of the surroundings is given by:

ΔS surroundings = ΔS energy + ΔS work

= (-13.4 J/K/mol) + (-0.008 J/K)

= -13.4 J/K/mol - 0.008 J/K

= -13.408 J/K/mol

This is the correct answer for the entropy change of the surroundings.

To answer your question about when to consider work, it is important to consider work whenever there is a change in volume or when there is a force acting on the system. In this case, the system is expanding against a constant pressure, so work must be considered. In other cases, such as an isothermal process with no change in volume, work may not need to be considered. It is important to carefully consider the process and determine if work needs to be taken into account in calculating entropy changes.
 

Related to Solve Thermodynamics Problem: Expansion Irreversibly and Isothermally at 27°C

1. What is thermodynamics?

Thermodynamics is a branch of physics that deals with the study of energy and its transformations, specifically the relationship between heat, work, and other forms of energy.

2. What is an irreversible expansion?

An irreversible expansion is a process in which the system undergoes a change in volume without being in equilibrium with its surroundings. This means that the expansion is not reversible and cannot be returned to its original state without a loss of energy.

3. What is an isothermal process?

An isothermal process is a process in which the temperature of a system remains constant throughout the process. This means that the system's internal energy remains constant and there is no heat transfer involved.

4. How do you solve a thermodynamics problem involving expansion irreversibly and isothermally at 27°C?

To solve this type of problem, you would need to use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. You would also need to consider the specific heat and volume of the system to calculate the change in internal energy and work done.

5. What is the significance of the temperature being specified at 27°C?

The specified temperature of 27°C is important because it allows you to determine the initial and final states of the system, which is necessary to calculate the change in internal energy and work done. It also ensures that the process remains isothermal, meaning the temperature remains constant throughout the expansion.

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