Calculate work in following process during A to B, in following graph. (Given (CV = 3R)

To find the work done during the process from state A to state B, we first analyze the parameters given in the P-V diagram:
At state A:
Pressure, $P_A=8\text{atm}$
Volume, $V_A=0.5\text{L}$

At state B:
Pressure, $P_B=1\text{atm}$
Volume, $V_B=4\text{L}$

Now, let us check the product of pressure and volume at both states A and B:
At state A: $P_A{V}_A=8\times 0.5=4\text{L}·\text{atm}$
At state B: $P_B{V}_B=1\times 4=4\text{L}·\text{atm}$

Since the product of pressure and volume remains constant ($PV=\text{constant}$), this indicates that the temperature of the system is constant throughout the process. Thus, the path from A to B represents a reversible isothermal expansion of an ideal gas.

According to the IUPAC convention in thermodynamics, the work done ($W$) during a reversible isothermal expansion is given by:
$W=-nRT\ln \left(\frac{V_B}{V_A}\right)$

Since $nRT=PV=4\text{L}·\text{atm}$, we can substitute this value into the equation:
$W=-(4\text{L}·\text{atm})\times \ln \left(\frac{4\text{L}}{0.5\text{L}}\right)$

Simplify the natural log term:
$W=-4\times \ln \left(8\right)\text{L}·\text{atm}$
Using the logarithmic property $\ln \left(x^y\right)=y\ln \left(x\right)$:
$W=-4\times \ln \left(2^3\right)\text{L}·\text{atm}$
$W=-4\times 3\ln \left(2\right)\text{L}·\text{atm}$
$W=-12\times \ln \left(2\right)\text{L}·\text{atm}$

Given that $\ln \left(2\right)\approx 0.693$:
$W\approx -12\times 0.693\text{L}·\text{atm}$
$W\approx -8.316\text{L}·\text{atm}$

Rounding to one decimal place, we obtain:
$W\approx -8.3\text{L}·\text{atm}$