Question

A sample of gas at temperature $T$ is adiabatically expanded to double its volume The work done by the gas in the process is (given, \(\gamma=\frac{3}{2}\))

• A. $W=T R[\sqrt{2}-2]$
• B. $W=\frac{R}{T}[2-\sqrt{2}]$
• C. $W=R T[2-\sqrt{2}]$
• D. $W=\frac{T}{R}[\sqrt{2}-2]$

Answer 1

The Correct Option is C

To solve the problem of finding the work done by the gas during adiabatic expansion, we need to use the principles of thermodynamics. Specifically, for an adiabatic process, we can use the following equations and steps:

In an adiabatic process, there is no heat exchange with the surroundings, and the process obeys the relation:

\(PV^{\gamma} = \text{constant}\)

Where \(\gamma\) (gamma) is the adiabatic index or heat capacity ratio, given as \(\frac{3}{2}\) in this problem.

The work done by a gas undergoing an adiabatic process can be calculated using the formula:

\(W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}\)

Initially, let the volume be \(V_1\). During adiabatic expansion, the volume doubles, so the final volume \(V_2 = 2V_1\).

Using the adiabatic condition:

\(P_1 V_1^{\gamma} = P_2 V_2^{\gamma}\)

Substitute \(V_2 = 2V_1\) and rearrange to find \(P_2\):

\(P_2 = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma} = P_1 \left(\frac{1}{2}\right)^{\frac{3}{2}}\)

So now, substituting \(P_2 = P_1 \cdot \frac{1}{2^{\frac{3}{2}}}\) and \(V_2 = 2V_1\) into the work formula:

\(W = \frac{P_1 V_1 - P_1 \cdot \frac{1}{2^{\frac{3}{2}}} \cdot 2V_1}{\frac{3}{2} - 1}\)

Calculate \(1 - \frac{1}{2^{\frac{1}{2}}}\):

\(W = \frac{P_1 V_1 \left(1 - \frac{1}{\sqrt{2}}\right)}{\frac{1}{2}}\)

\(W = 2P_1 V_1 \left(1 - \frac{1}{\sqrt{2}}\right)\)

Using the ideal gas equation, \(P_1 V_1 = nRT\), the work done can be expressed as:

\(W = 2nRT \left(1 - \frac{1}{\sqrt{2}}\right) = RT \left(2 - \sqrt{2}\right)\)

Thus, the correct answer is:

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

Therefore, the correct option is: $W = RT[2 - \sqrt{2}]$.