<MultiModal>![In a "U"-shaped reactor, frictionless and massless heat-conducting pistons $\mathbf{1}$ and $\mathbf{2}$ are fixed by pins at the center and right side, respectively, dividing the adiabatic reactor into a left section $\mathbf{A}$ and a right section $\mathbf{B}$](/media/uploads/bdbe5e99-bd7d-4a45-abaa-50273223e783.png)</MultiModal>

In the adiabatic reactor shown in the figure, the following reactions take place respectively:

$$
\mathrm{N_2O_4}_{(g)}\rightleftharpoons \mathrm{2NO_2}_{(g)},K_p^\circ=0.141
$$

Initially, both sections have reached chemical equilibrium, with equal volumes and total pressures of $2p^\circ$ and $p^\circ$ ($p^\circ = 101325 \mathrm{Pa}$), respectively, at a temperature of $298\mathrm {K} $.  
If the pin on piston $\mathbf{1}$ is removed, what will be the volume ratio of sections $\mathbf{A}$ and $\mathbf{B}$ when the piston stabilizes in its final position?
<MultiModal>![一个“匚”形反应器中，无摩擦、无质量的导热活塞$\mathbf{1}$和$\mathbf{2}$分别被销钉固定在中间和右侧，将绝热反应器分成左侧的$\mathbf{A}$部分和右侧$\mathbf{B}$部分](/media/uploads/bdbe5e99-bd7d-4a45-abaa-50273223e783.png)</MultiModal>

在如图所示绝热反应器中, 分别进行以下反应：

$$
\mathrm{N_2O_4}_{(g)}\rightleftharpoons \mathrm{2NO_2}_{(g)},K_p^\circ=0.141
$$

起始时，两部分均已达到化学平衡，体积相等，总压分别为 $2p^\circ$ 和 $p^\circ$（$p^\circ = 101325 \mathrm{Pa}$），温度均为 $298\mathrm {K} $。
若将活塞$\mathbf{1}$上的销钉拔掉，最终活塞位置稳定时，$\mathbf{A}$、$\mathbf{B}$两部分的体积比为多少？
2.096
2.000
1.048
1.500
4.192
None of the above answers are correct
2.096
2.000
1.048
1.500
4.192
以上答案均不正确
In the initial state, consider the equilibrium when any portion of $\mathrm{N_2O_4}$ is introduced. Assuming the total pressure, $\mathrm{N_2O_4}$ partial pressure, and $\mathrm{NO_2}$ partial pressure are $p$, $p_\mathrm{N_2O_4}$, and $p_\mathrm{NO_2}$, respectively, and the degree of dissociation is $\alpha$, the following relationships hold at equilibrium:  
$$
p_\mathrm{NO_2} = 2\alpha p \\
p_\mathrm{N_2O_4} = (1-\alpha) p \\
K_p^\circ = \frac{p_\mathrm{NO_2}^2}{p_\mathrm{N_2O_4}p^\circ} = \frac{4\alpha^2}{1-\alpha}\frac{p}{p^\circ}
$$

<Checkpoint>[Dissociation degree equation $K_p^\circ = \frac{p_\mathrm{NO_2}^2}{p_\mathrm{N_2O_4}p^\circ} = \frac{4\alpha^2}{1-\alpha}\frac{p}{p^\circ}$](2)</Checkpoint>

Given $K_p^\circ$ and the total pressure $p$, the degree of dissociation $\alpha$ can be calculated, and subsequently, the partial pressures of gases on sides $\mathrm{A}$ and $\mathrm{B}$ are:  
$$
p_\mathrm{A}(\mathrm{N_2O_4}) = 155.5\ \text{kPa},\ p_\mathrm{A}(\mathrm{NO_2})=47.5\ \text{kPa}\\
p_\mathrm{B}(\mathrm{N_2O_4}) = 69.7\ \text{kPa},\ p_\mathrm{B}(\mathrm{NO_2})=31.6\ \text{kPa}
$$

<Checkpoint>[$$ p_\mathrm{A}(\mathrm{N_2O_4}) = 155.5\ \text{kPa},\ p_\mathrm{A}(\mathrm{NO_2})=47.5\ \text{kPa}\\ p_\mathrm{B}(\mathrm{N_2O_4}) = 69.7\ \text{kPa},\ p_\mathrm{B}(\mathrm{NO_2})=31.6\ \text{kPa} $$](2)</Checkpoint>

According to the ideal gas law $PV=nRT$, the total nitrogen ratio between $\mathrm{A}$ and $\mathrm{B}$ before removing the pin is:  
$$
\frac{n_A(\mathrm{NO_2})}{n_B(\mathrm{NO_2})}=\frac{2p_\mathrm{A}(\mathrm{N_2O_4})+p_\mathrm{A}(\mathrm{NO_2})}{2p_\mathrm{B}(\mathrm{N_2O_4})+p_\mathrm{B}(\mathrm{NO_2})}=2.096
$$

<Checkpoint>[The total nitrogen content ratio between $\mathbf{A}$ and $\mathbf{B}$ before removing the pin is 2.096](1)</Checkpoint>

The piston can conduct heat, so the temperatures on both sides are the same, and the equilibrium constants are identical.  

<Checkpoint>[The piston can conduct heat, so the temperatures on both sides are the same, and the equilibrium constants are identical](1)</Checkpoint>

After removing the pin on piston $\mathrm{1}$, when the piston stabilizes, the total pressures on sides $\mathrm{A}$ and $\mathrm{B}$ are equal. Therefore, according to the above equations, the degrees of dissociation on both sides are the same, and the partial pressures of the same gas are equal.  

<Checkpoint>[After removing the pin, the degrees of dissociation on both sides are the same, and the partial pressures of the same gas are equal](1)</Checkpoint>

Based on the ideal gas law, partial pressure relationships, and proportionality, the volume ratio after removing the pin can be related to the total nitrogen content, yielding the result:  
$$
\frac{V_\text{A}^t}{V_\text{B}^t}=\frac{n^t_\text{A,\ total}}{n^t_\text{B, total}}\\
=\frac{n^t_\mathrm{A,\ N_2O_4}}{n^t_\mathrm{B,\ N_2O_4}}=\frac{n^t_\mathrm{A,\ NO_2}}{n^t_\mathrm{B,\ NO_2}}\\
=\frac{2n^t_\mathrm{A,\ N_2O_4}+n^t_\mathrm{A,\ NO_2}}{2n^t_\mathrm{B,\ N_2O_4}+n^t_\mathrm{B,\ NO_2}}\\
=\frac{n_A(\mathrm{NO_2})}{n_B(\mathrm{NO_2})}=2.096
$$

<Checkpoint>[After removing the pin, the volume ratio between $\mathbf{A}$ and $\mathbf{B}$ equals the nitrogen content ratio, which is 2.096](2)</Checkpoint>
初始状态时, 考虑任意部分放入$\mathrm{N_2O_4}$时的平衡. 假设总压, $\mathrm{N_2O_4}$分压和$\mathrm{NO_2}$分压为$p$, $p_\mathrm{N_2O_4}$和$p_\mathrm{NO_2}$, 解离度为$\alpha$, 反应平衡时满足
$$
p_\mathrm{NO_2} = 2\alpha p \\
p_\mathrm{N_2O_4} = (1-\alpha) p \\
K_p^\circ = \frac{p_\mathrm{NO_2}^2}{p_\mathrm{N_2O_4}p^\circ} = \frac{4\alpha^2}{1-\alpha}\frac{p}{p^\circ}
$$

<Checkpoint>[解离度方程$K_p^\circ = \frac{p_\mathrm{NO_2}^2}{p_\mathrm{N_2O_4}p^\circ} = \frac{4\alpha^2}{1-\alpha}\frac{p}{p^\circ}$](2)</Checkpoint>

由于$K_p^\circ$和总压$p$已知, 可算出解离度$\alpha$, 进而算出$\mathrm{A}$和$\mathrm{B}$两侧各气体分压:
$$
p_\mathrm{A}(\mathrm{N_2O_4}) = 155.5\ \text{kPa},\ p_\mathrm{A}(\mathrm{NO_2})=47.5\ \text{kPa}\\
p_\mathrm{B}(\mathrm{N_2O_4}) = 69.7\ \text{kPa},\ p_\mathrm{B}(\mathrm{NO_2})=31.6\ \text{kPa}
$$

<Checkpoint>[$$ p_\mathrm{A}(\mathrm{N_2O_4}) = 155.5\ \text{kPa},\ p_\mathrm{A}(\mathrm{NO_2})=47.5\ \text{kPa}\\ p_\mathrm{B}(\mathrm{N_2O_4}) = 69.7\ \text{kPa},\ p_\mathrm{B}(\mathrm{NO_2})=31.6\ \text{kPa} $$](2)</Checkpoint>


根据理想气体状态方程$PV=nRT$, 可计算得拔去销钉前,$\mathrm{A}$与$\mathrm{B}$总氮比例为
$$
\frac{n_A(\mathrm{NO_2})}{n_B(\mathrm{NO_2})}=\frac{2p_\mathrm{A}(\mathrm{N_2O_4})+p_\mathrm{A}(\mathrm{NO_2})}{2p_\mathrm{B}(\mathrm{N_2O_4})+p_\mathrm{B}(\mathrm{NO_2})}=2.096
$$

<Checkpoint>[拔去销钉前A和B总氮含量比例为2.096](1)</Checkpoint>

活塞可以传热, 因此两侧温度一致, 平衡常数一致; 

<Checkpoint>[活塞可以传热, 因此两侧温度一致, 平衡常数一致](1)</Checkpoint>

拔掉活塞$\mathrm{1}$上的销钉后, 活塞稳定时, $\mathrm{A}$与$\mathrm{B}$两侧总压相等; 因此, 根据上述公式, 两侧解离度一致, 同种气体分压相等.

<Checkpoint>[拔去销钉后两侧解离度一致，同种气体分压相等](1)</Checkpoint>

根据理想气体状态方程, 分压关系和比例的性质, 可得拔去销钉后体积比与总氮含量的关系, 进而计算出结果
$$
\frac{V_\text{A}^t}{V_\text{B}^t}=\frac{n^t_\text{A,\ total}}{n^t_\text{B, total}}\\
=\frac{n^t_\mathrm{A,\ N_2O_4}}{n^t_\mathrm{B,\ N_2O_4}}=\frac{n^t_\mathrm{A,\ NO_2}}{n^t_\mathrm{B,\ NO_2}}\\
=\frac{2n^t_\mathrm{A,\ N_2O_4}+n^t_\mathrm{A,\ NO_2}}{2n^t_\mathrm{B,\ N_2O_4}+n^t_\mathrm{B,\ NO_2}}\\
=\frac{n_A(\mathrm{NO_2})}{n_B(\mathrm{NO_2})}=2.096
$$

<Checkpoint>[拔去销钉后$\mathbf{A}$、$\mathbf{B}$两部分的体积比即为两侧氮含量比，为2.096](2)</Checkpoint>