Question

analyse the change in entropy accompanying isothermal expansion of 5 moles of an ideal gas at 330 K , until its volume has increased 6 times​

Answer 1

The entropy change is $35.13 \frac{J}{K}$.

Explanation:

• From the first law of thermodynamics, we can write Δu=q+w;

• As the give process is isothermal, the temperature is constant through out the change, therefore internal energy of the system is constant.
• ⇒Δu=0
• therefore, $q+w=0$

⇒$w=-qrev$

⇒Δs be the entropy change and entropy is $\frac{-qrev}{T}$

⇒Δs= $\frac{w}{T}$ and work done in an isothermal process is $nRTlog(\frac{V2}{V1})$------------------(1)

Given volume has increased 6 times, ⇒$V2-V1=6V1$

therefore $V2=7V1$

and also the temperature through out the change is 330K and number of moles of gas is n=5. we know universal gas constant R=8.314 J/mol.K

substituting these values in equation(1), we get Δs=$\frac{nRTlog(\frac{V2}{V1} )}{T} = nRlog\frac{V2}{V1}$

therefore, Δs= $5*8.314*log(\frac{7}{1} )= 35.13$$\frac{J}{K}$