An ideal gas (0.5 mol), initially at 2 bar pressure, is compressed at a constant temperature of 600 K in two steps: first, against a constant external pressure of P bar (2 < P < 8), and then against constant external pressure of 8 bar. At each step, the compression is stopped only when the pressure of the gas becomes equal to the external pressure. The total work done on the gas in these steps is W. Considering all possible values of P (2 < P < 8) and taking the gas constant as R (in J K⁻¹mol⁻¹), the minimum value of |W| (in J) is equal to:

The correct option is 600R.

Let us solve the problem step-by-step by analyzing the work done during the two-step irreversible compression of an ideal gas.

1. Understanding the Given Data:
Number of moles of the gas, $n=0.5\text{mol}$
Initial pressure of the gas, $P_0=2\text{bar}$
Temperature of the gas (constant), $T=600\text{K}$
Intermediate external pressure, $P_1=P\text{bar}$ where $2<P<8$
Final external pressure, $P_2=8\text{bar}$

2. Work Done in Step 1:
The gas is compressed against a constant external pressure $P_1=P$ until its pressure reaches $P$.
Using the ideal gas equation $PV=nRT$, the initial volume $V_0$ and volume after Step 1 ($V_1$) are:

$V_0=\frac{nRT}{2}$

$V_1=\frac{nRT}{P}$

The work done on the gas in Step 1 ($W_1$) is given by:

$W_1=-P_{\text{ext}}(V_1-V_0)=-P\left(\frac{nRT}{P}-\frac{nRT}{2}\right)=nRT\left(\frac{P}{2}-1\right)$

3. Work Done in Step 2:
Next, the gas is compressed against a constant external pressure $P_2=8\text{bar}$ until its pressure reaches $8\text{bar}$.
The final volume $V_2$ is:

$V_2=\frac{nRT}{8}$

The work done on the gas in Step 2 ($W_2$) is given by:

$W_2=-8(V_2-V_1)=-8\left(\frac{nRT}{8}-\frac{nRT}{P}\right)=nRT\left(\frac{8}{P}-1\right)$

4. Total Work Done ($W$):
Summing the work done in both steps:

$W=W_1+W_2=nRT\left(\frac{P}{2}-1+\frac{8}{P}-1\right)=nRT\left(\frac{P}{2}+\frac{8}{P}-2\right)$

Since $n=0.5$ and $T=600\text{K}$:

$nRT=0.5\times R\times 600=300R$

Therefore, the expression for total work done becomes:

$W=300R\left(\frac{P}{2}+\frac{8}{P}-2\right)$

5. Finding the Minimum Value of $\left|W\right|$:
Since work is done on the gas during compression, $W>0$, so $\left|W\right|=W$.
To minimize $W$ with respect to $P$, we apply the AM-GM inequality to the terms $\frac{P}{2}$ and $\frac{8}{P}$:

$\frac{\frac{P}{2}+\frac{8}{P}}{2}\ge \sqrt{\frac{P}{2}\times \frac{8}{P}}=\sqrt{4}=2$

$\frac{P}{2}+\frac{8}{P}\ge 4$

Equality holds (yielding the minimum value) when:

$\frac{P}{2}=\frac{8}{P}\Rightarrow P^2=16\Rightarrow P=4\text{bar}$

Substituting $P=4\text{bar}$ back into the total work formula:

$W_{\text{min}}=300R\left(\frac{4}{2}+\frac{8}{4}-2\right)=300R(2+2-2)=300R\times 2=600R$

Thus, the minimum value of $\left|W\right|$ is 600R.