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Entropy is given with equation:

$S=k_b*\lnΩ$

Where $S$ - entropy, $k_b$ - Boltzmann's constant and $Ω$ - number of possible microstates.

What quantities is the number of microstates dependent on?

Suppose I have a 1 litre container, isolated from outside environment, filled with 0.5 litres of gas at 50 °K and 0.5 litres of the same gas at 100 °K, both at the same pressure p. Eventually this system will reach an equilibrium at some temperature and the entropy of the system will have increased, because heat was transferred. The volume, number of molecules and internal energy is still the same for the system, how could the entropy have increased? What has increased in the system for it to have more possible microstates?

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The volume, number of molecules and internal energy is still the same for the system

But entropy is only a function of these state variables when the system is in equilibrium, whereas your system is initially out of equilibrium so you can't conclude that the entropy should remain constant.

What has increased in the system for it to have more possible microstates?

There are many more microstates in which the temperature is uniform than where the temperature is (say) 50 K in one half and 100 K in the other half.

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The entropy of the whole system is maximized in equilibrium. There are more microstates that lead to the macro state "temperature and volume of the gas take their equilibrium values", than there are microstates corresponding to the macro state "temperature and volume taking the values you started with."

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The multiplicity of a combined system is the product of the multiplicities of the partial systems. Thermodynamic equilibrium occurs when total Ω is at is maximum, that is the most likely state. Temperatures are equal then. Then the fractional changes of multiplicities Ω for a small change in internal energy are equal for both partial systems.

When temperatures are not equal, random processes transfer energy from one part of the system to the other part, the multiplicity of the first system goes down and but that of the second system goes up by a larger percentage. This will increase the product of their multiplicities.

In your example, at 100 K the thermodynamic $\beta = \frac{1}{kT} = \frac{1}{\Omega}\ \frac{{\rm d}\Omega}{{\rm d}E}$ is about 12 % per meV and 50 K is about 24 % per meV. When one milli-eV in thermal energy is transferred from the 100 K part to the 50 K part, the multiplicity of the whole system goes up by about by a factor $0.88 \times 1.24 = 1.09$.

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