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| | | ### Statement |
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| | − | $5.8.15.$ [Insert the problem statement] |
| | + | $5.8.15.$ The piston initially divides the cylindrical vessel into two equal parts, in which there is an ideal gas of the same mass with the same temperature. Is it a real process in which, as the piston moves, the temperature of one part increases twice, and the other part decreases twice? The heat capacity of the piston and cylinder can be ignored, the system is isolated. |
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| | | ### Solution |
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| | − | I consider this problem to be a cornerstone of this section. In my opinion, it is this specific problem that fully reveals the probabilistic thermodynamic approach in statistical physics. Some might find the solution overly detailed, but I would like to demonstrate it in the style of Sivukhin's methodology. Any macrostate of a system characterized by pressure, volume, and temperature is realized by a vast number of microstates—that is, by a specific set of coordinates and momenta of all molecules. The number of such microstates represents the thermodynamic probability \(W\). According to the second law of thermodynamics, an isolated system can spontaneously transition only from a less probable state to a more probable one. That is, for any real process, the ratio of the final probability of the system to the initial one must be: |
| | + | I consider this problem to be a cornerstone of this section. In my opinion, it is this specific problem that fully reveals the probabilistic thermodynamic approach in statistical physics. Some might find the solution overly detailed, but I would like to demonstrate it in the style of Sivukhin's methodology. |
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| | + | Any macrostate of a system characterized by pressure, volume, and temperature is realized by a vast number of microstates—that is, by a specific set of coordinates and momenta of all molecules. The number of such microstates represents the thermodynamic probability $W$. |
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| | + | According to the second law of thermodynamics, an isolated system can spontaneously transition only from a less probable state to a more probable one. That is, for any real process, the ratio of the final probability of the system to the initial one must be: |
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| | + | $$\frac{W_{f}}{W_{i}}\ge 1$$ |
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| | + | Since the molecules of an ideal gas do not interact with each other, the spatial distribution of the particles and their velocity (energy) distribution are independent events. Therefore, the total thermodynamic probability of a single part of the gas can be represented as the product of the spatial component $W_{V}$ and the temperature component $W_{T}$: |
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| | + | Let a gas consisting of $N$ molecules be contained in a vessel of total volume $V_{tot}$, and we consider the probability of the gas occupying a certain part of this volume, $V$. The probability $p$ for a single randomly chosen molecule to be inside the selected volume is: |
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| | + | $$p=\frac{V}{V_{tot}}$$ |
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| | + | Since the molecules of an ideal gas move independently of one another, the probability that all $N$ molecules will simultaneously be within the volume $V$ is found using the multiplication rule for probabilities: |
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| | + | $$P=p^{N}=\left( \frac{V}{V_{tot}} \right)^{N}$$ |
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| | + | The thermodynamic probability $W_{V}$ is directly proportional to this mathematical probability. Since the total volume of the vessel does not change during the process, it follows that: |
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| | + | $$W_{V}\sim V^{N} \tag{1}$$ |
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| | + | The temperature component indicates how many ways a fixed internal energy $U$ can be distributed among all the molecules. Let each molecule have $i$ degrees of freedom. In total, a system of $N$ molecules will have $M=iN$ degrees of freedom. |
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| | + | The microstate of the system in terms of energy is specified by a set of momenta for each degree of freedom: $\left(p_{1},p_{2},...,p_{M} \right)$. The total internal energy of the gas $U$ is: |
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| | + | $$U=\frac{p_{1}^{2}}{2m}+\frac{p_{2}^{2}}{2m}+...+\frac{p_{M}^{2}}{2m}\Longrightarrow 2mU=p_{1}^{2}+p_{2}^{2}+...+p_{M}^{2}$$ |
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| | + | Geometrically, this equation defines a sphere in an $M$-dimensional momentum space, the radius of which is $R=\sqrt{2mU}$. The number of microstates $W_{T}$ is proportional to the surface area of this multidimensional sphere. From geometry, it is known that the area of an $M$-dimensional sphere is proportional to its radius raised to the power of $M-1$. Since $M\sim 10^{23}$, the unit can be neglected: |
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| | + | $$W_{T}\sim R^{M}=(\sqrt{2mU})^{M}\sim U^{\frac{M}{2}}$$ |
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| | + | Since the internal energy of an ideal gas is directly proportional to its absolute temperature, it follows that: |
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| | + | $$W_{T}\sim T^{\frac{M}{2}}=T^{\frac{i}{2}N} \tag{2}$$ |
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| | + | Thus, the total thermodynamic probability of the macrostate is: |
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| | + | $$W\sim V^{N}\cdot T^{\frac{i}{2}N} \tag{3}$$ |
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| | + | In our problem, the cylinder is divided by a piston into two isolated parts, each containing the same mass of gas, which implies an identical number of molecules $N$. Since the subsystems are independent, the total thermodynamic probability of the entire system equals the product of the probabilities of its halves: |
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| | + | $$W_{sys}=W_{1}\cdot W_{2}$$ |
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| | + | Initially, both parts occupy the same volume $V_{0}$ and have the same temperature $T_{0}$: |
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| | + | $$W_{i}=W_{1}\cdot W_{2}\sim \left( V_{0}^{N}\cdot T_{0}^{\frac{i}{2}N} \right)\cdot\left( V_{0}^{N}\cdot T_{0}^{\frac{i}{2}N} \right)=V_{0}^{2N}\cdot T_{0}^{iN} \tag{4}$$ |
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| | + | In the final state, the temperature of the first part has doubled, $T_{1}=2T_{0}$, while that of the second part has halved, $T_{2}=\frac{T_{0}}{2}$. |
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| | + | Suppose that during the movement of the piston, the first part occupies a volume $V_{1}=x\cdot V_{0}$, where $x$ is the volume change coefficient. Since the total volume of the cylinder is fixed and equals $2V_{0}$, the second part is forced to occupy the volume: $V_{2}=2V_{0}-V_{1}=(2-x)V_{0}$. |
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| | + | The probability of the system in the final state is: |
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| | + | $$W_{f}\sim \left[ (xV_{0})^{N}\cdot (2T_{0})^{\frac{i}{2}N} \right]\cdot\left[ \left( (2-x)V_{0} \right)^{N}\cdot \left( \frac{T_{0}}{2} \right)^{\frac{i}{2}N} \right]$$ |
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| | + | From here: |
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| | + | $$W_{f}\sim V_{0}^{2N}\cdot T_{0}^{iN}\cdot\left[ x(2-x) \right]^{N} \tag{5}$$ |
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| | + | The ratio of the final probability of the system to the initial one is: |
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| | + | $$\frac{W_{f}}{W_{i}}=\frac{V_{0}^{2N}\cdot T_{0}^{iN}\cdot\left[ x(2-x) \right]^{N}}{V_{0}^{2N}\cdot T_{0}^{iN}}=\left[ x(2-x) \right]^{N} \tag{6}$$ |
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| | + | The function $f(x)=x(2-x)=1-(1-x)^{2}$ is a downward-opening parabola. The maximum possible value of the function is 1 and is achieved only at $x=1$, when the piston remains exactly in the middle, meaning no process occurs. For any displacement of the piston $x\neq 1$, the value of the function is strictly less than unity, $f(x)\lt 1$. Since the number of molecules in the gas is macroscopically huge, raising a number less than unity to the power of $N$ yields a quantity that is practically zero: |
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| | + | $$\frac{W_{f}}{W_{i}}\to 0\Longrightarrow W_{f}\ll W_{i} \tag{7}$$ |
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| | + | The probability of the final state of the system for any shift of the piston turns out to be statistically negligible compared to the initial one. According to the second law of thermodynamics, an isolated system cannot spontaneously transition into a state with a lower thermodynamic probability. Therefore, the described spontaneous process is absolutely impossible. |
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| | | #### Answer |
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| | − | [Insert a concise answer or boxed result] |
| | + | Unrealistic |
