A sealed flask with a capacity of $$2 \text{ dm}^3$$ contains $$11 \text{ g}$$ of propane gas. The flask is so weak that it will burst if the pressure becomes $$2 \text{ MPa}$$. The minimum temperature at which the flask will burst is ______ °C. [Nearest integer]
(Given: $$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$$. Atomic masses of C and H are $$12u$$ and $$1u$$ respectively.) (Assume that propane behaves as an ideal gas.)

We need to find the minimum temperature at which a sealed flask will burst.

Volume = $$2 \text{ dm}^3 = 2 \times 10^{-3} \text{ m}^3$$

Mass of propane = 11 g

Burst pressure = 2 MPa = $$2 \times 10^6 \text{ Pa}$$

$$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$$

Calculate moles of propane

Molecular formula of propane: $$C_3H_8$$

Molar mass = $$3 \times 12 + 8 \times 1 = 44 \text{ g/mol}$$

$$n = \frac{11}{44} = 0.25 \text{ mol}$$

Apply the ideal gas equation

$$PV = nRT$$

$$T = \frac{PV}{nR}$$

Substitute values

$$T = \frac{2 \times 10^6 \times 2 \times 10^{-3}}{0.25 \times 8.3}$$

$$T = \frac{4000}{2.075}$$

$$T = 1927.7 \text{ K}$$

Convert to Celsius

$$T = 1927.7 - 273 = 1654.7 \approx 1655 \text{ °C}$$

The answer is 1655.