Question

A sample of 1 mole gas at temperature T is adiabatically expanded to double its volume. If adiabatic constant for the gas is \(\gamma=\frac{3}{2}\) , then the work done by the gas in the process is:

• A. \(RT[2-\sqrt2]\)
• B. \(\frac{R}{T}[2-\sqrt2]\)
• C. \(RT[2+\sqrt2]\)
• D. \(\frac{T}{R}[2-\sqrt2]\)

Answer 1

The Correct Option is A

To find the work done by the gas during adiabatic expansion, we will use the following principle of adiabatic processes: For adiabatic processes, the formula that relates initial and final states is given by:

\[ PV^\gamma = \text{constant} \]

where \( \gamma \) is the adiabatic constant, here being \( \frac{3}{2} \).

We will consider the initial and final states and the principle that in adiabatic processes, \( PV^\gamma = \text{constant} \).

Suppose the initial volume is \( V \) and the initial pressure is \( P \). After expansion, the final volume is \( 2V \) (since the volume is doubled). Let the final pressure be \( P' \).

By using the adiabatic condition:

\[ P \cdot V^\gamma = P' \cdot (2V)^\gamma \]

Simplifying further,

\[ P = P' \cdot 2^\gamma \]

Substitute \( \gamma = \frac{3}{2} \):

\[ P = P' \cdot 2^{\frac{3}{2}} \]

The work done \( W \) in an adiabatic process is given by:

\[ W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} \]

Substituting \( \gamma \) and rearranging terms, we get:

\[ W = \frac{P \cdot V - P \cdot 2^{-\frac{1}{2}} \cdot 2V}{\frac{3}{2} - 1} \]

Simplifying the expression:

\[ W = \frac{P \cdot V (1 - \sqrt{2})}{\frac{1}{2}} = 2P \cdot V (1 - \sqrt{2}) \]

Using the ideal gas law, \( PV = nRT \), where \( n = 1 \) mole:

\[ W = 2 \cdot (RT) \cdot (1 - \sqrt{2}) \]

Thus, the work done by the gas in the process is:

\[ W = RT[2 - \sqrt{2}] \]

Therefore, the correct option is: \(RT[2 - \sqrt{2}]\).

Answer 2

Step 1: Adiabatic condition The adiabatic condition is given by:

\(TV^{\gamma-1} = \text{constant.}\)

For the initial and final states:

\(T(V)^{\frac{3}{2}-1} = T_f(2V)^{\frac{3}{2}-1}.\)

Simplify:

\(TV^{\frac{1}{2}} = T_f(2V)^{\frac{1}{2}},\)

\(TV\sqrt{V} = T_f\sqrt{2V}.\)

Cancel \(\sqrt{V}\):

\(T = T_f\sqrt{2}.\)

Solve for \(T_f\):

\(T_f = \frac{T}{\sqrt{2}}.\)

Step 2: Work done in adiabatic expansion The work done in an adiabatic process is given by:

\(W.D. = \frac{nR}{1-\gamma}\left[T_f - T\right].\)

Substitute \(T_f = \frac{T}{\sqrt{2}}, \gamma = \frac{3}{2},\) and \(n = 1\):

\(W.D. = \frac{R}{1-\frac{3}{2}}\left[\frac{T}{\sqrt{2}} - T\right].\)

Simplify:

\(W.D. = \frac{R}{-\frac{1}{2}}\left[\frac{T}{\sqrt{2}} - T\right],\)

\(W.D. = -2R\left[\frac{T}{\sqrt{2}} - T\right].\)

Factorize:

\(W.D. = 2RT\left[1 - \frac{1}{\sqrt{2}}\right].\)

Simplify further:

\(W.D. = RT\left[2 - \sqrt{2}\right].\)

Final Answer: \(RT\left[2 - \sqrt{2}\right].\)