Question

One mole of an ideal gas of volume \(V\) and temperature \(T\) is allowed to expand adiabatically to volume \(2V\) while doing no external work. The universal gas constant is \(R\). What is the pressure of the gas after expansion?

• A. \(\frac{RT}{2V}\)
• B. \(\frac{RT}{4V}\)
• C. \(\frac{RT}{V}\)
• D. \(\frac{2RT}{V}\)

Hint:

Do not confuse "adiabatic free expansion" with "reversible adiabatic expansion".
In reversible adiabatic expansion, \(P V^{\gamma} = \text{constant}\).
But in free expansion, no work is done, temperature is constant, and the process is highly irreversible.
Simply apply the ideal gas law with \(T = \text{constant}\).

Answer 1

The Correct Option is A

To solve the problem, we need to find the pressure of the gas after an adiabatic expansion where no external work is done. First, let's understand the key concepts involved:

Concept of Adiabatic Process: In an adiabatic process, there is no heat exchange between the system and surroundings. For an ideal gas undergoing adiabatic expansion, the relationship between pressure, volume, and temperature can be given by the equation:

\(PV^{\gamma} = \text{constant}\) where \(\gamma\) is the ratio of specific heats, \(\gamma = \frac{C_p}{C_v}\).

However, in this particular problem, the gas does no external work during the expansion process. This implies that the internal energy change is zero, which only occurs for free expansion of an ideal gas where the final temperature \(T'\) remains equal to the initial temperature \(T\).

Given:

• Initial Volume, \(V_i = V\)
• Final Volume, \(V_f = 2V\)
• Initial Temperature, \(T_i = T\)
• Final Temperature, \(T_f = T\) (Since no external work is done, and the process is adiabatic)

We need to determine the pressure at this final state. Using the ideal gas law:

\(PV = nRT\)

For initial state: \(P_i V_i = RT\)

For final state: \(P_f V_f = RT_f\)

Since \(T_f = T\), the equation becomes:

\(P_f \times 2V = RT\)

Rearrange to find \(P_f\):

\(P_f = \frac{RT}{2V}\)

This matches the answer: \(\frac{RT}{2V}\).

Conclusion: The correct answer is that the pressure of the gas after expansion is \(\frac{RT}{2V}\), which corresponds to the first option.