In the given figure, when the membrane separating the gas from the evacuated region is ruptured, the gas expands freely and irreversibly. The container is thermally insulated from its surroundings. The entropy change of the gas (in $J / K$ ) is. (Take $R = \frac{25}{3} J/ (mol \cdot K), ln 2 = 0.7$ )

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TestHub · Accepted Answer

In free expansion the gas does no work and the container is insulated, so the temperature of the ideal gas stays constant.

The entropy change is computed along a reversible isothermal path joining the same two states:

$Δ S = n R ln \frac{V _{2}}{V _{1}}$

From the figure, 1 mole of gas expands from volume H to H + 3H, i.e. 4H:

$Δ S = 1 \times \frac{25}{3} \times ln \frac{4 H}{H}$

$= \frac{25}{3} \times ln 4$

$Δ S = \frac{25}{3} \times 2 ln 2 = \frac{25}{3} \times 2 \times 0.7$

$= \frac{35}{3} \approx 11.67 J/K$

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