1. Given data

• Ratio of specific heats: $\gamma = 1.5 = \dfrac{3}{2}$
• Initial volume: $V_1 = 1200\,\text{cm}^3 = 1200 \times 10^{-6}\,\text{m}^3 = 1.2 \times 10^{-3}\,\text{m}^3$
• Final volume: $V_2 = 300\,\text{cm}^3 = 300 \times 10^{-6}\,\text{m}^3 = 3.0 \times 10^{-4}\,\text{m}^3$
• Initial pressure: $P_1 = 200\,\text{kPa} = 2.0 \times 10^5\,\text{Pa}$

We need the absolute value of work done by the gas during an adiabatic compression.

2. Use adiabatic relation

For an adiabatic process,

$P_1 V_1^\gamma = P_2 V_2^\gamma$

So,

$P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma$

Now,

$\frac{V_1}{V_2} = \frac{1200}{300} = 4$

Thus,

$P_2 = 2.0 \times 10^5 \times 4^{1.5}$

Since,

$4^{1.5} = 4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$

Therefore,

$P_2 = 2.0 \times 10^5 \times 8 = 1.6 \times 10^6\,\text{Pa}$

3. Work done in adiabatic process

For an adiabatic process,

$W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$

Now calculate:

$P_1V_1 = 2.0 \times 10^5 \times 1.2 \times 10^{-3} = 240\,\text{J}$

$P_2V_2 = 1.6 \times 10^6 \times 3.0 \times 10^{-4} = 480\,\text{J}$

Hence,

$W = \frac{240 - 480}{1.5 - 1} = \frac{-240}{0.5} = -480\,\text{J}$

The negative sign indicates that during compression, work done by the gas is negative.

So the absolute value is

$|W| = 480\,\text{J}$

4. Final answer

$\boxed{480}$

5. Comparison with stored correct answer

Stored correct answer = $480$

Our derived answer also = $480$, so they agree.