We’re given an ideal gas initially at $$T_i = 0^\circ \mathrm{C} = 273\,\mathrm{K}.$$ The gas is suddenly compressed to one-fourth of its initial volume, and we are told that

$$\gamma = \frac{c_p}{c_v} = \frac{3}{2}\,.$$

For an adiabatic process (one in which there is no heat exchange), if the process were reversible, the relation between the temperature and volume would be

$$T\, V^{\gamma-1} = \text{constant}\,.$$

Even though the compression is “sudden” (and hence irreversible), the final equilibrium state of the gas is uniquely determined by its internal energy (which is a state function). That allows us to use the adiabatic relation between the initial and final states.

Here are the steps:

Write the adiabatic relation between the initial and final states:

$$T_i\, V_i^{\gamma-1} = T_f\, V_f^{\gamma-1}\,.$$

Since the gas is compressed to one-fourth of its volume, we have

$$V_f = \frac{1}{4} V_i\,.$$

Substitute this into the adiabatic relation:

$$273\,V_i^{\gamma-1} = T_f \left(\frac{1}{4} V_i\right)^{\gamma-1}\,.$$

With $$\gamma = \frac{3}{2},$$

$$\gamma - 1 = \frac{3}{2} - 1 = \frac{1}{2}\,.$$

So the relation becomes:

$$273\,V_i^{1/2} = T_f \left(\frac{1}{4}\right)^{1/2} V_i^{1/2}\,.$$

Cancel the common factor $$V_i^{1/2}$$ (provided it is nonzero) and simplify:

$$273 = T_f \left(\frac{1}{2}\right)\,.$$

Now solve for the final temperature $$T_f$$:

$$T_f = 273 \times 2 = 546\,\mathrm{K}\,.$$

The change in temperature is then:

$$\Delta T = T_f - T_i = 546 - 273 = 273\,\mathrm{K}\,.$$

So, the change in temperature due to the thermodynamic process is $$\boxed{273\,\mathrm{K}}.$$