\(2\) mole of an ideal gas at \(27°C\) temperature is expanded reversibly from \(2 L\) to \(20 L\). Find entropy change \((R = 2 cal/mol/K)\).
Answer & explanation
Correct answer: option 4
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