# Calculating Entropy Change for an Ideal Gas Expansion

• chemochem
In summary, the conversation is about calculating entropy for an isothermal and irreversible process. The formula used is ##\Delta S = \int_i^f \frac{dQ_{rev}}{T}## and the values for the system and surroundings are calculated separately and added together to find the total entropy change for the universe. The correct value for the surroundings should be negative and determined by the heat transferred in the irreversible path.
chemochem
MENTOR NOTE: NO TEMPLATE BECAUSE SUMITTED TO WRONG FORUM.

3.1) A quantity of 0.10 mol of an ideal gas A initially at 22.2 degrees C is expanded from 0.200 dm3 to 2.42 dm3 . Calculate the values of work (w), heat (q), internal energy change (delta U), entropy change of the system (deltaSsys), entropy change of the surroundings (deltaSsurr), and total entropy change (delta S univ) if the process is carried out isothermally and irreversibly against an external pressure of 1.00 atm.

The question is how do I calculate entropy? I tried something, and this is what I got:
delta S system=2,0728 JK-1
delta S surroundings=0.76 JK-1
delta S universe=2,8328 JK-1

Can you please check this out?

Last edited by a moderator:
chemochem said:
3.1) A quantity of 0.10 mol of an ideal gas A initially at 22.2 degrees C is expanded from 0.200 dm3 to 2.42 dm3 . Calculate the values of work (w), heat (q), internal energy change (delta U), entropy change of the system (deltaSsys), entropy change of the surroundings (deltaSsurr), and total entropy change (delta S univ) if the process is carried out isothermally and irreversibly against an external pressure of 1.00 atm.

The question is how do I calculate entropy? I tried something, and this is what I got:
delta S system=2,0728 JK-1
delta S surroundings=0.76 JK-1
delta S universe=2,8328 JK-1

Can you please check this out?
Hi Chemochem. Welcome to PF!

It is difficult to check what you have done since you have not explained what you did.

To calculate the change in entropy, you need to calculate ##\Delta S = \int_i^f \frac{dQ_{rev}}{T}## where dQrev is the heat flow during a reversible process from the initial to final states. (Note: For irreversible processes, the reversible path between initial and final states of the system will not be the same path as the reversible path between the initial and final states of the surroundings). In this case, what would be a reversible path between the initial and final states for the system? For the surroundings? Calculate ##\Delta S_{sys} \text{ and } \Delta S_{surr}## and add them together to the ##\Delta S_{univ}##.

AM

Well, your delta S for the surroundings is definitely wrong. It should be negative. Show us the details please.

I used delta S=nRln(Vf/Vi) for delta S system and then I found a reversible path and calculated the value as being 0.76 JK-1 for delta S surroundings (it turned out to be -w/T). I may have forgotten to write the minus sign, I dont know.

chemochem said:
I used delta S=nRln(Vf/Vi) for delta S system and then I found a reversible path and calculated the value as being 0.76 JK-1 for delta S surroundings (it turned out to be -w/T). I may have forgotten to write the minus sign, I dont know.
The change in entropy for the surroundings should be -Q/T for the actual irreversible path.

Then if its isothermal, dU is equal to zero, so the value for Q is just -w, right?

chemochem said:
Then if its isothermal, dU is equal to zero, so the value for Q is just -w, right?
Yes. Do you understand why, for the surroundings, ##\Delta S## is determined by the heat transferred in the irreversible path?

Yes, I used w because it had the same value and then forgot the sign, so I got the wrong result. Thank you!

## 1. What is entropy change for an ideal gas expansion?

Entropy change for an ideal gas expansion is a measure of the disorder or randomness of the gas particles during the expansion process. It is a thermodynamic property that describes the change in the distribution of energy and particles in a system.

## 2. How is entropy change calculated for an ideal gas expansion?

The entropy change for an ideal gas expansion can be calculated using the formula ΔS = nRln(V2/V1), where ΔS is the change in entropy, n is the number of moles, R is the gas constant, and V1 and V2 are the initial and final volumes of the gas, respectively.

## 3. What are the units for entropy change in an ideal gas expansion?

The units for entropy change in an ideal gas expansion are joules per kelvin (J/K) or calories per kelvin (cal/K). These units represent the change in energy per unit temperature during the expansion process.

## 4. How does temperature affect the entropy change in an ideal gas expansion?

The temperature of the gas affects the entropy change in an ideal gas expansion. As the temperature increases, the entropy change also increases. This is because at higher temperatures, the gas particles have more kinetic energy and therefore more disorder, leading to a larger change in entropy.

## 5. What are the assumptions made when calculating entropy change for an ideal gas expansion?

When calculating entropy change for an ideal gas expansion, it is assumed that the gas is expanding isothermally (at a constant temperature) and reversibly (slowly enough to maintain equilibrium). It is also assumed that the gas behaves ideally, meaning there are no intermolecular forces between particles and no volume occupied by the particles themselves.

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