Entropy Change of an expanding gas

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SUMMARY

The change in entropy for two moles of an ideal gas undergoing a reversible isothermal expansion from 1.97×10-2 m3 to 4.82×10-2 m3 at a temperature of 20.0°C is calculated using the formula ΔS = nR ln(V2/V1). The correct calculation yields an entropy change of 14.9 J/K, correcting the initial erroneous value of -138.20 J/K. This highlights the importance of accurate mathematical computation in thermodynamic calculations.

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cmcc3119
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Two moles of an ideal gas undergo a reversible isothermal expansion from 1.97×10−2 m^3 to 4.82×10−2 m^3 at a temperature of 20.0 C.

What is the change in entropy of the gas?

WORKING:

Using \DeltaS = nRIn(V2/V1)

= 2(8.31*ln(4.82*10^-2/1.97*10^-2)

ANS = -138.20 J/K

Is this correct? I can only enter it once on my website and I want to make sure this is right first time :)
 
Last edited:
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Hi cmcc3119,

I think you made a math error in your calculation; I don't think that number follows from the line above it.
 
Yep I figured out the right anwer which is 14.9

I don't know what I meant with that massive negative figure! Thanks for you help anyway.
 

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