An LPG cylinder contains gas at a pressure of 300 kPa at 27°C. The cylinder can withstand the pressure of $$1.2 \times 10^6$$ Pa. The room in which the cylinder is kept catches fire. The minimum temperature at which the bursting of cylinder will take place is _________ °C. (Nearest integer)

We have an LPG cylinder that contains gas initially at a pressure of $$P_1 = 300 \,\text{kPa} = 3 \times 10^5 \,\text{Pa}$$ and an initial temperature of $$t_1 = 27^\circ\text{C}$$. To work with gas-law equations we must always convert Celsius to Kelvin, because the Kelvin scale starts from absolute zero. The conversion formula is

$$T (\text{K}) = t (^\circ\text{C}) + 273.$$

Applying this conversion, the initial absolute temperature is

$$T_1 = 27 + 273 = 300 \,\text{K}.$$

The cylinder can safely withstand a maximum pressure of $$P_2 = 1.2 \times 10^6 \,\text{Pa}.$$ We are asked to find the temperature $$t_2$$ (in °C) at which the cylinder will reach this pressure and burst.

An LPG cylinder is a rigid container; therefore its volume remains constant during heating. For a fixed amount of an ideal gas kept at constant volume, pressure is directly proportional to absolute temperature. This relation is expressed by Gay-Lussac’s (or the Isochoric) gas law:

$$\frac{P}{T} = \text{constant} \qquad\Longrightarrow\qquad \frac{P_1}{T_1} = \frac{P_2}{T_2}.$$

Rearranging this formula to solve for the unknown absolute temperature $$T_2$$, we write

$$T_2 = \frac{P_2 \, T_1}{P_1}.$$

Now we substitute each known value carefully:

$$T_2 = \frac{1.2 \times 10^6 \,\text{Pa} \times 300 \,\text{K}}{3 \times 10^5 \,\text{Pa}}.$$

First, notice that the pascal (Pa) units cancel out. Next, perform the arithmetic with the numbers and powers of ten:

$$T_2 = \frac{1.2 \times 10^6}{3 \times 10^5} \times 300 \,\text{K}.$$

The fraction of the coefficients is $$\frac{1.2}{3} = 0.4$$, and the fraction of the powers of ten is $$\frac{10^6}{10^5} = 10^{6-5} = 10^1 = 10$$. Therefore

$$T_2 = 0.4 \times 10 \times 300 \,\text{K}.$$

Multiplying step by step, $$0.4 \times 10 = 4$$, and then $$4 \times 300 = 1200.$$ Hence

$$T_2 = 1200 \,\text{K}.$$

We now convert this absolute temperature back to the Celsius scale by reversing the earlier formula:

$$t_2 (^\circ\text{C}) = T_2 (\text{K}) - 273.$$

Substituting $$T_2 = 1200 \,\text{K}$$, we get

$$t_2 = 1200 - 273 = 927^\circ\text{C}.$$

This temperature is the minimum room temperature at which the cylinder pressure will reach $$1.2 \times 10^6 \,\text{Pa}$$ and consequently burst.

So, the answer is $$927^\circ\text{C}.$$