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Calculate the entropy change when 5 moles of a monatomic ideal gas undergoes an isothermal expansion from an initial volume of 8 liters to a final volume of 40 liters. The initial temperature is 300 \mathrm{~K}.
 

Option: 1

15 \mathrm{~J} / \mathrm{K}


 


Option: 2

13.38 \mathrm{~J} / \mathrm{K}
 


Option: 3

14.8 \mathrm{~J} / \mathrm{K}
 


Option: 4

19 \mathrm{~J} / \mathrm{K}


Answers (1)

Step 1: Use the adiabatic process to relate the initial and final conditions for an ideal gas.
\mathrm{ P_1 V_1^\gamma=P_2 V_2^\gamma }

Given: Number of moles, \mathrm{n=5} moles Initial volume, \mathrm{V_1=8} liters Final volume, \mathrm{V_2=40} liters Initial temperature, \mathrm{T_1=300 \mathrm{~K}}

Specific heat ratio (adiabatic index), \mathrm{\gamma=\frac{C_p}{C_v}}, typically for monatomic gases \mathrm{\gamma \approx 1.4}

Step 2: Calculate the initial and final pressures using the adiabatic process equation.

\mathrm{ P_1=\frac{n R T_1}{V_1} }

\mathrm{ P_2=\frac{n R T_2}{V_2}}

Step 3: Use the relation between pressure and volume in an adiabatic process to find the temperature at the final state.

\mathrm{ T_2=\frac{P_2 V_2}{n R} }

Step 4: Calculate the change in entropy \mathrm{ (\Delta S) } for an isothermal process using the ideal gas equation.
\mathrm{ \Delta S=R \ln \left(\frac{V_2}{V_1}\right) }
Step 5: Substitute the values and calculate the entropy change.

\mathrm{ \Delta S=R \ln \left(\frac{40}{8}\right) \\ }

\mathrm{ \Delta S=13.38 J / K }

So,correct option is 2.

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Kshitij

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