1. Given data

• Initial temperature: $T_1 = 0^\circ C = 273\,\text{K}$
• Final volume: gas is compressed to one fourth of initial volume $V_2 = \frac{V_1}{4}$
• Ratio of specific heats: $\gamma = \frac{C_p}{C_v} = \frac{3}{2}$

Because the gas is compressed suddenly, the process is taken as adiabatic.

2. Use adiabatic relation

For an ideal gas in an adiabatic process, $T V^{\gamma-1} = \text{constant}$

So, $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$

Hence, $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$

Substitute $V_2 = V_1/4$ and $\gamma - 1 = \frac{3}{2}-1 = \frac{1}{2}$:

$\frac{T_2}{T_1} = \left(\frac{V_1}{V_1/4}\right)^{1/2} = 4^{1/2} = 2$

Therefore, $T_2 = 2T_1 = 2 \times 273 = 546\,\text{K}$

3. Find change in temperature

$\Delta T = T_2 - T_1 = 546 - 273 = 273\,\text{K}$

4. Final answer

The change in temperature is: $\boxed{273\,\text{K}}$

5. Comparison with stored answer

Stored correct answer: $273$

Our derived answer is also $273$, so they agree.