Concept Explanation: Irreversible Adiabatic Expansion of an Ideal Gas

This problem involves an ideal gas expanding irreversibly against zero external pressure in a thermally insulated vessel. Let's analyze the situation step by step.

Step 1: Identify the Type of Process

The vessel is thermally insulated, meaning no heat exchange occurs with the surroundings. Therefore, $q=0$. This is an adiabatic process.

Step 2: Analyze the Work Done

The gas expands against zero external pressure ($P_{ext}=0$). The work done by the gas in an irreversible expansion is given by:

$w=-P_{ext}\mathrm{\Delta }V$

Since $P_{ext}=0$, the work done is:

$w=0$

Step 3: Apply the First Law of Thermodynamics

The first law states: $\mathrm{\Delta }U=q+w$

We have already established that q=0 and $w=0$. Therefore:

$\mathrm{\Delta }U=0+0=0$

For an ideal gas, the internal energy $U$ depends only on temperature. If $\mathrm{\Delta }U=0$, then the temperature must remain constant.

$\mathrm{\Delta }T=0\Rightarrow T_2=T_1$

Step 4: Apply the Ideal Gas Law

Since the temperature is constant ($T_2=T_1$), we can apply Boyle's Law for the initial and final states.

Initial state: $P_1{V}_1=nR{T}_1$

Final state: $P_2{V}_2=nR{T}_2$

Because $T_2=T_1$, the right-hand sides of both equations are equal. Therefore:

$P_1{V}_1=P_2{V}_2$

Step 5: Evaluate the Other Option

The option $P_2{V}_2^{\gamma }=P_1{V}_1^{\gamma }$ is the condition for a reversible adiabatic process. Our process is irreversible and adiabatic, but it is also isothermal ($\mathrm{\Delta }T=0$). The relation $P{V}^{\gamma }=\text{constant}$ is not valid here because it applies only to adiabatic processes where temperature changes.

Final Answer

For this expansion, the following statements are correct:

• $q=0$ (The process is adiabatic)
• $T_2=T_1$ (The temperature remains constant)
• $P_2{V}_2=P_1{V}_1$ (Boyle's Law holds for this isothermal change)

The statement $P_2{V}_2^{\gamma }=P_1{V}_1^{\gamma }$ is incorrect for this process.

Related Topics & Formulae

Key Concepts

• First Law of Thermodynamics: $\mathrm{\Delta }U=q+w$
• Ideal Gas Law: $PV=nRT$
• Internal Energy of an Ideal Gas: $\mathrm{\Delta }U=n{C}_v\mathrm{\Delta }T$ (depends only on temperature change)
• Work in Irreversible Expansion: $w=-P_{ext}\mathrm{\Delta }V$

Adiabatic Processes

An adiabatic process is one where no heat is exchanged ($q=0$).

• Reversible Adiabatic Process: Follows the relation $T{V}^{\gamma -1}=\text{constant}$ or $P{V}^{\gamma }=\text{constant}$, where $\gamma =\frac{C_p}{C_v}$.
• Irreversible Adiabatic Process: The above relations do not hold. The process must be analyzed using the first law of thermodynamics and the specific conditions (like $P_{ext}=0$ in this case).

Why This Process is Special

This is a unique case of an adiabatic process that is also isothermal. It only happens when an ideal gas expands against zero pressure, resulting in zero work. Since no work is done and no heat is exchanged, there is no change in internal energy, and hence, no change in temperature.