\(2\) mole of an ideal gas at \(27°C\) temperature is expanded reversibly from \(2 L\) to \(20 L\). Find entropy change \((R = 2 cal/mol/K)\).

9.2

The correct answer is option 4. 9.2.

To find the entropy change (\(\Delta S\)) for the reversible isothermal expansion of an ideal gas, we can use the following formula:

\(\Delta S = nR \ln \frac{V_f}{V_i}\)

where:

\( n \) is the number of moles,

\( R \) is the gas constant,

\( V_f \) is the final volume,

\( V_i \) is the initial volume.

Given:

\( n = 2 \) moles,

\( T = 27^\circ \text{C} \) (though temperature remains constant and doesn't directly affect the calculation of entropy change for an isothermal process),

\( V_i = 2 \, \text{L} \),

\( V_f = 20 \, \text{L} \),

\( R = 2 \, \text{cal/mol/K} \).

Now, plug the values into the formula:

\(\Delta S = 2 \times 2 \, \text{cal/mol/K} \times \ln \frac{20}{2} \)

\(\Delta S = 4 \, \text{cal/K} \times \ln 10 \)

We need the natural logarithm of 10 (\(\ln 10\)):

\(\ln 10 \approx 2.3026 \)

Now, calculate \(\Delta S\):

\(\Delta S = 4 \, \text{cal/K} \times 2.3026 \)

\(\Delta S \approx 9.2104 \, \text{cal/K} \)

Rounding to one decimal place, we get:

\(\Delta S \approx 9.2 \, \text{cal/K}\)

Therefore, the entropy change is: (4) 9.2