Question

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

• A. $RT\left[\sqrt{2} - 2\right]$
• B. $RT\left[1 - 2\sqrt{2}\right]$
• C. $RT\left[2\sqrt{2} - 1\right]$
• D. $RT\left[2 - \sqrt{2}\right]$

Answer 1

Answer: (D)

$RT\left[2 - \sqrt{2}\right]$

Answer 2

For an adiabatic process, the work done $W$ is given by:
$W = \frac{nR\Delta T}{1-\gamma}.$

1. Using the Adiabatic Condition:
Since the process is adiabatic, $TV^{\gamma-1} = \text{constant}$. Let the initial temperature be $T$ and the final temperature be $T_f$ when the volume is doubled. Thus,
$TV^{\gamma-1} = T_f(2V)^{\gamma-1}.$

2. Calculate $T_f$:
Simplifying, we get:
$T_f = T \left(\frac{1}{2}\right)^{\frac{\gamma-1}{\gamma}} = T \left(\frac{1}{2}\right)^{\frac{1}{2}} = \frac{T}{\sqrt{2}}.$

3. Calculate the Work Done:
Substitute into the work formula:
$W = \frac{R(T - T_f)}{1 - \frac{3}{2}} = \frac{R \left( T - \frac{T}{\sqrt{2}} \right)}{-\frac{1}{2}}.$ Simplifying further:
$W = 2RT\frac{\left(\sqrt{2} - 1\right)}{\sqrt{2}} = RT(2 - \sqrt{2}).$ Answer: $RT(2 - \sqrt{2})$