Calculating Entropy Change of Ideal Gas

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To calculate the entropy change of 3 moles of an ideal gas during isothermal compression from 5V to V at 30°C, the process involves understanding that for isothermal conditions, the internal energy change (dU) is zero. The relationship for an ideal gas under isothermal conditions is expressed as PV = constant, not PV^gamma, which applies to adiabatic processes. The entropy change can be determined using the formula for entropy change (dS) in terms of volume change, integrating the expression derived from the first law of thermodynamics. The correct approach involves recognizing the nature of the process and applying the appropriate equations for isothermal conditions. This method will yield the desired entropy change for the gas.
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I have found the following question on a previous exam paper whilst I've been doing my revision, and have a few questions about how to set up my answer.

Question:
Calculate the entropy change of 3 moles of an ideal gas when they are isothermally compressed (T = 30° C) from a volume 5V to a volume V.

My initial thought was to use PV = nRT, but because both n and T stay constant I then started thinking that I could just make P1V1 = P2V2.

However from my notes I have that PV^gamma = constant which confused me as there is no mention of a value for gamma in the question.

From here I would use the forumla W=-integralP(V)dV and then use the S=W/T formula.

Is this the correct way to go with this question?

Thanks for any feedback
 
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It's simpler than that. Try to answer the following questions in order.

1.) For an isothermal process, what is dU?
2.) Can you find an expression for dU in terms of dS and dV? (since after all you were given 2 volumes and you were asked for the change in entropy).
3.) Can you integrate the expression found in 2.)?

That should get you there.
 
PV^1 = constant for an isothermal process (treating it as an ideal gas ...)
PV^gamma = constant for an adiabatic process.
 
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