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Chemistry · THERMODYNAMICS
हिन्दीIn an experiment, \( 1\text{ mol} \) of an ideal gas at \( 300\text{ K} \) expands from \( 2\text{ L} \) to \( 20\text{ L} \). In case (i), it expands isothermally and reversibly. In case (ii), it expands isothermally against a constant external pressure of \( 1\text{ atm} \). What is the ratio of the magnitude of work done in case (i) to that in case (ii)? (Take \( R = 0.082\text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} \), \( \ln 10 = 2.303 \))
Correct answer · Option 4
\( 3.15 \)
Explanation
|w_rev| = 2.303 nRT log(20/2) = 2.303 * 1 * 0.082 * 300 * 1 = 56.65 L atm. |w_irrev| = p_ex * Delta V = 1 * (20 - 2) = 18 L atm. Ratio = 56.65 / 18 = 3.147 ~ 3.15.
