Initially a gas of diatomic molecules is contained in a cylinder of volume V1 at a pressure p1 and t

Q: Why does the number of moles change during dissociation?

When a diatomic molecule dissociates, the bond between the two atoms breaks, resulting in two separate monoatomic atoms. This means one mole of diatomic gas produces two moles of monoatomic gas upon complete dissociation, increasing the total particle count.

Q: How did we get $1.25\mu_0$ for the final moles?

Initially, we had $\mu_0$ moles. 25% dissociated, leaving $0.75\mu_0$ diatomic moles. The 25% that dissociated ($0.25\mu_0$) formed twice as many monoatomic moles ($2 \times 0.25\mu_0 = 0.5\mu_0$). Adding them together: $0.75\mu_0 + 0.5\mu_0 = 1.25\mu_0$.

Q: Why can we use the ideal gas law for a mixture of diatomic and monoatomic gases?

The ideal gas law $pV = nRT$ depends only on the total number of moles of gas particles, regardless of whether they are diatomic or monoatomic, as long as they behave ideally and don't interact chemically.

Q: What happens to the gas constant $R$?

The universal gas constant $R$ remains the same for all ideal gases and mixtures. It cancels out perfectly when we take the ratio of the two state equations.

Q: Could we solve this using the combined gas law $p_1 V_1 / T_1 = p_2 V_2 / T_2$?

No, that combined gas law only applies when the amount of gas (number of moles) remains constant. Since dissociation changes the number of moles, we must use the full ideal gas equation $pV = nRT$ for both states.

Question

Initially a gas of diatomic molecules is contained in a cylinder of volume $V_1$ at a pressure $p_1$ and temperature $250\mathrm{K}$. Assuming that $25\%$ of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature $2000\mathrm{K}$, when contained in a volume $2V_1$ is given by $p_2$. The ratio $p_2/p_1$ is ...........

Sigma Percentile · Accepted Answer

### The Dynamics of Dissociation: A Journey Through the Ideal Gas Law

Imagine a closed cylinder filled with a diatomic gas. This is our starting point, a perfectly stable system governed by the elegant rules of thermodynamics. Let's assume the initial number of moles of this diatomic gas is $\mu_0$. The pressure is $p_1$, the volume is $V_1$, and the temperature is a cool $250\mathrm{K}$.

By applying the ideal gas equation, $pV = nRT$, we can perfectly describe this initial state.

$$p_1 V_1 = \mu_0 R (250)$$

This simple equation is our anchor. It ties together all the macroscopic properties of the gas before any dramatic changes occur.

### The Catalyst of Change

Now, the system undergoes a massive transformation. The volume is forcibly doubled to $2V_1$, and the temperature is cranked up to a blistering $2000\mathrm{K}$. But the most critical change happens at the microscopic level: **dissociation**.

The intense heat causes $25\%$ of the diatomic molecules to break their bonds. This is where many students make a critical error. When one diatomic molecule (like $O_2$) breaks apart, it doesn't just disappear; it forms **two** separate monoatomic atoms (like $2O$). This fundamental conservation of atoms means the total number of particles—and therefore the total number of moles—will increase.

Let's calculate the new number of moles, $\mu_2$. The remaining, intact diatomic molecules make up $75\%$ of the original amount, which is $0.75\mu_0$. The $25\%$ that dissociated ($0.25\mu_0$) yields twice as many monoatomic moles, giving us $2 	imes 0.25\mu_0 = 0.5\mu_0$.

Adding these together gives us the total moles in the final state:

$$\mu_2 = 0.75\mu_0 + 0.5\mu_0 = 1.25\mu_0$$

### The Grand Synthesis

Armed with our new mole count, we can write the ideal gas equation for the final state. The new pressure is $p_2$, the volume is $2V_1$, and the temperature is $2000\mathrm{K}$.

$$p_2 (2V_1) = (1.25\mu_0) R (2000)$$

To find the ratio of the final pressure to the initial pressure, $\frac{p_2}{p_1}$, we simply divide our final state equation by our initial state equation. This is a beautiful mathematical maneuver because all the constants and initial variables—like $V_1$, $\mu_0$, and the universal gas constant $R$—will gracefully cancel out.

$$\frac{p_2 (2V_1)}{p_1 V_1} = \frac{1.25\mu_0 R (2000)}{\mu_0 R (250)}$$

Simplifying the right side, we divide $2000$ by $250$ to get $8$.

$$2 \left(\frac{p_2}{p_1}ight) = 1.25 	imes 8$$

Multiplying $1.25$ by $8$ yields exactly $10$. Finally, dividing both sides by $2$ reveals our answer.

$$\frac{p_2}{p_1} = 5$$

The pressure in the final state is exactly five times the initial pressure. This problem beautifully illustrates how macroscopic changes in volume and temperature, combined with microscopic changes like molecular dissociation, collectively dictate the final state of a gas.

Enter Numerical Value:

Visualized Solution

$Initial State Setup$

$Ideal Gas Law for State 1$

$Final State Setup$

$Dissociation of Molecules$

$Calculating Final Moles$

$Ideal Gas Law for State 2$

$Taking the Ratio$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Kinetic Theory of Gases

The Dynamics of Dissociation

The Catalyst of Change

The Grand Synthesis

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ideal Gas Law for State 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final State Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dissociation of Molecules

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Final Moles

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ideal Gas Law for State 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Taking the Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward