A sample of gas with $$\gamma = 1.5$$ is taken through an adiabatic process in which the volume is compressed from 1200 cm$$^3$$ to 300 cm$$^3$$. If the initial pressure is 200 kPa. The absolute value of the workdone by the gas in the process = _________ J.

For an adiabatic process of an ideal gas we always have the relation

$$P\,V^{\gamma}= \text{constant}.$$

The work done by the gas (positive for expansion, negative for compression) in a reversible adiabatic process is given by the formula

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

Here the given data are

$$\gamma = 1.5, \quad V_1 = 1200\ \text{cm}^3, \quad V_2 = 300\ \text{cm}^3, \quad P_1 = 200\ \text{kPa}.$$

First we convert the volumes into SI units (m$$^3$$):

$$V_1 = 1200\ \text{cm}^3 = 1200\times10^{-6}\ \text{m}^3 = 1.2\times10^{-3}\ \text{m}^3,$$

$$V_2 = 300\ \text{cm}^3 = 300\times10^{-6}\ \text{m}^3 = 3.0\times10^{-4}\ \text{m}^3.$$

Using $$P_1V_1^{\gamma}=P_2V_2^{\gamma},$$ we find the final pressure:

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

The volume ratio is

$$\frac{V_1}{V_2}= \frac{1.2\times10^{-3}}{3.0\times10^{-4}} = 4,$$ so

$$P_2 = 200\ \text{kPa}\;\bigl(4\bigr)^{1.5}.$$

Because $$4^{1.5}=4^{3/2}=(4^1)(4^{1/2})=4\times2=8,$$ we obtain

$$P_2 = 200\ \text{kPa}\times 8 = 1600\ \text{kPa}.$$

Next we calculate the products $$P_1V_1$$ and $$P_2V_2,$$ remembering that $$1\ \text{kPa}=10^{3}\ \text{Pa}$$ and $$1\ \text{Pa}\cdot\text{m}^3=1\ \text{J}.$$

$$P_1V_1 = 200\ \text{kPa}\times1.2\times10^{-3}\ \text{m}^3 = 200\times10^{3}\ \text{Pa}\times1.2\times10^{-3}\ \text{m}^3$$

$$\phantom{P_1V_1} = 200\times1.2\times(10^{3}\times10^{-3})\ \text{J} = 240\ \text{J}.$$

$$P_2V_2 = 1600\ \text{kPa}\times3.0\times10^{-4}\ \text{m}^3 = 1600\times10^{3}\ \text{Pa}\times3.0\times10^{-4}\ \text{m}^3$$

$$\phantom{P_2V_2} = 1600\times3.0\times(10^{3}\times10^{-4})\ \text{J} = 480\ \text{J}.$$

Substituting these values into the work formula that is convenient for magnitude,

$$W = \left|\frac{P_1V_1-P_2V_2}{\gamma-1}\right|,$$

we have

$$P_1V_1-P_2V_2 = 240\ \text{J}-480\ \text{J} = -240\ \text{J},$$

and since $$\gamma-1 = 1.5-1 = 0.5,$$

$$W = \left|\frac{-240\ \text{J}}{0.5}\right| = \left|-480\ \text{J}\right| = 480\ \text{J}.$$

The negative sign merely indicated that the gas was compressed; the question asks for the absolute value, which is 480 J.

So, the answer is $$480\ \text{J}.$$