Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4% when heated by 1°C. Its initial temperature must be:

For a fixed amount of ideal gas in a closed rigid vessel, the volume remains constant. According to Gay-Lussac’s law (also called the pressure law), at constant volume the pressure $$P$$ of an ideal gas is directly proportional to its absolute temperature $$T$$.
Mathematically, $$\frac{P}{T} = \text{constant} \quad\Longrightarrow\quad \frac{\Delta P}{P} = \frac{\Delta T}{T}$$

The problem states: the pressure rises by $$0.4\%$$ upon heating the gas by $$1^{\circ}\text{C}$$. Writing these changes in symbols,

$$\frac{\Delta P}{P} = 0.4\% = \frac{0.4}{100} = 0.004$$

Since a change of $$1^{\circ}\text{C}$$ equals a change of $$1\text{ K}$$ on the absolute scale,

$$\Delta T = 1\text{ K}$$

Using the proportionality relation:

$$\frac{\Delta T}{T} = \frac{\Delta P}{P}$$

$$\Rightarrow \quad \frac{1\text{ K}}{T} = 0.004$$

Solving for the initial absolute temperature $$T$$:

$$T = \frac{1}{0.004} = 250 \text{ K}$$

Hence the initial temperature of the gas is $$250\text{ K}$$. Converting to the Celsius scale, $$250\text{ K} - 273 \approx -23^{\circ}\text{C}$$, but none of the given options list this negative Celsius value; instead, option C directly offers $$250\text{ K}$$.

Therefore, the correct choice is Option C (250 K).