To determine the total work done when a perfect gas undergoes a transformation from point 1 to point 4, we can analyze each step of the process:

Given Data

Moles of gas, $n = 0.1 \, \text{mol}$

Specific heat at constant volume, $\overline{\mathrm{C}}_v=1.50 \mathrm{R}$

Gas constant, $\mathrm{R}=0.082 \, \mathrm{L} \, \mathrm{atm} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$

Work Done Calculation

Step 1: From Point 1 to Point 2 ($W_{1 \rightarrow 2}$)

Since this is an isochoric process (constant volume), the work done, $W_{1 \rightarrow 2}$, is zero.

$ \mathrm{W}_{1 \rightarrow 2} = 0 $

Step 2: From Point 2 to Point 3 ($W_{2 \rightarrow 3}$)

This step is isobaric (constant pressure), and the work done can be calculated as:

$ \mathrm{W}_{2 \rightarrow 3} = -P\Delta V = -P(V_3 - V_2) $

Given:

$P = 3 \, \text{atm}$

Change in volume, $\Delta V = V_3 - V_2 = 2 - 1 = 1 \, \text{L}$

$ \mathrm{W}_{2 \rightarrow 3} = -3 \, \text{atm} \times 1 \, \text{L} = -3 \, \text{L atm} $

Converting to Joules (using $1 \, \text{L atm} = 101.3 \, \text{J}$):

$ \mathrm{W}_{2 \rightarrow 3} = -3 \times 101.3 \, \text{J} = -304 \, \text{J} $

Step 3: From Point 3 to Point 4 ($W_{3 \rightarrow 4}$)

This is another isochoric process, so the work done, $W_{3 \rightarrow 4}$, is zero.

$ \mathrm{W}_{3 \rightarrow 4} = 0 $

Total Work Done

The total work done from point 1 to point 4 is the sum of work done in each step:

$ \text{Total work done} = W_{1 \rightarrow 2} + W_{2 \rightarrow 3} + W_{3 \rightarrow 4} = 0 + (-304) + 0 = -304 \, \text{J} $

Thus, the total work done during this transformation is $-304 \, \text{J}$.