Consider a container of volume $ 5.0$ L that is divided into two compartments of equal size. In the left compartment there is nitrogen at $1.0$ $atm$ and $25 °C$; in the right compartment there is hydrogen at the same temperature and pressure. What will happen when the partition is removed?

$A) $The entropy decreases, and the free energy decreases.
$B)$ The entropy increases, and the free energy decreases.
$C) $The entropy increases, and the free energy increases.
$D) $The entropy decreases, and the free energy increases.

Logic tells that upon removing the partition, randomness increases and hence entropy increases. I am confused about free energy. First law of thermodynamics has to be applied , I think. But I can't seem to get the right direction.


A Spontaneous process is characterized by an increase in the total entropy (for both system and surroundings).

Spontaneous processes are characterized by a decrease in free energy (analogous to the decrease in gravitational potential energy occurring for a ball rolling downhill).


We know that free energy of a process is given by:

$$ \Delta G_\mathrm{mix} = \Delta H_\mathrm{mix} - T \Delta S_\mathrm{mix} $$

The molar entropy of mixing is given by:

$$\Delta S_\mathrm{mix} = -R(x_\ce{A}\ln(x_\ce{A}) + x_\ce{B}\ln(x_\ce{B}))$$

For an ideal gas: $\Delta H_\mathrm{mix} = 0$ giving free energy as:

$$\Delta G_\mathrm{mix} = -T\Delta S_\mathrm{mix} = RT(x_\ce{A}\ln(x_\ce{A}) + x_\ce{B}\ln(x_\ce{B}))$$

since $x_\ce{A}$ and $x_\ce{B}$ are mole fractions ($x_i \le 1$), the natural log will always yield a negative number. Therefore the entropy will be positive and the free energy will be negative.


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