A perfect gas (0.1 mol) having $$\bar{C}_v = 1.50 R$$ (independent of temperature) undergoes the transformation from point 1 to point 4 as shown in the P-V diagram. If each step is reversible, the total work done (w) while going from point 1 to point 4 is $$(-) \text{____}$$ J (nearest integer).
[Given: $$R = 0.082$$ L atm $$K^{-1} mol^{-1}$$]

The total work done during the process can be obtained by calculating the work done in each step of the given path on the P-V diagram.

The expression for pressure-volume work is

$$W=-\int P_{\text{ext}},dV.$$

Since the process occurs in three stages,

$$1 \rightarrow 2,\quad 2 \rightarrow 3,\quad 3 \rightarrow 4,$$

the total work is

$$W_{\text{total}}=W_{1\rightarrow2}+W_{2\rightarrow3}+W_{3\rightarrow4}.$$

The given values of the number of moles ($0.1$ mol) and molar heat capacity

$$\bar{C}_v=1.50R$$

are not required for calculating pressure-volume work, since work depends only on the pressure and volume changes.

For the process from 1 to 2:

• The process is isochoric.
• The volume remains constant at

$$1000\ \text{cm}^3.$$

Since

$$\Delta V=0,$$

the work done is

$$W_{1\rightarrow2}=0.$$

For the process from 2 to 3:

• The process is isobaric.
• The pressure remains constant at

$$3.00\ \text{atm}.$$

The volume changes from

$$1000\ \text{cm}^3=1\ \text{L}$$

to

$$2000\ \text{cm}^3=2\ \text{L}.$$

Therefore,

$$\Delta V=2-1=1\ \text{L}.$$

Hence,

$$W_{2\rightarrow3}=-P\Delta V$$

$$=-3.00\times1$$

$$=-3.00\ \text{L atm}.$$

For the process from 3 to 4:

• The process is again isochoric.
• The volume remains constant at

$$2000\ \text{cm}^3.$$

Therefore,

$$\Delta V=0,$$

and

$$W_{3\rightarrow4}=0.$$

Adding the contributions from all three steps,

$$W_{\text{total}}=0-3.00+0=-3.00\ \text{L atm}.$$

Using the conversion

$$1\ \text{L atm}=101.325\ \text{J},$$

the total work is

$$W_{\text{total}}=-3.00\times101.325$$

$$=-303.975\ \text{J}.$$

Rounding to the nearest integer,

$$W_{\text{total}}=-304\ \text{J}.$$

Since the question is of the form

$$(-)\ \underline{\hspace{1cm}}\ \text{J},$$

the value to be filled in the blank is

304.