Knowledge of the probability of the state of a system in molecular physics makes it possible to predict the future behavior of this system.
You can implement an unlikely event. For example, in one half of the vessel, divided into two equal parts by a wall, there is gas. The wall is removed very quickly. A gas state is realized, the probability of which is 2−N, where N is the number of gas particles in the vessel. This result can be obtained by solving problem 5.8.1 c. At subsequent points in time, other states will occur in the vessel. The system won't return to its initial state — its probability is too small! Consequently, an irreversible transition to new, more probable states will occur, and the molecules will fill the entire space of the vessel.
This example shows that knowing the probability of the state of a new system is very useful.
And how to calculate the state probability for other systems? Wouldn't the calculations be too complicated? It turns out — no, not really. Calculate how many times the probability of one state is greater than the probability of another, as follows. If we give a system that is in state 1 at temperature T the amount of heat Q, then it will pass to another state, the probability of which is e^(Q/kT) times greater than the probability of state 1 (k is the Boltzmann constant).
Here is an example of how such calculations are performed. Find out how many times the probability of a state in which all N gas molecules are in one half of the vessel (state 1) is less than the probability of a state in which the molecules fill evenly the entire space of the vessel (state 2). We will block the vessel in which the molecules are located with a movable piston. When the piston is moved to the left by ∆x, the gas performs work ∆A and cools down. To maintain the temperature of the gas constant, we must give the gas molecules the amount of heat ∆Q = ∆A (to make up for the energy losses in the gas). Therefore, when moving the piston to the left, we will transfer an amount of heat Q = A to the gas at temperature T. For isothermal expansion, A = νRTln(V_k/V_h), where ν is the amount of gas (in moles), R = kN_A is the gas constant, N_A is the Avogadro constant, V_h is the initial volume, and V_k is the final gas volume.
In our case, N = νN_A, V_k/V_h = 2. Therefore, the formula can be rewritten as A = NkTln2. Hence, the probability of state 2 is e^(A/kT) = e^(Nln2) = 2^N times greater than the probability of state 1. We obtained the solution of problem