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EEJ MAIN Chemistry QUESTION #7513
Question 1

An LPG cylinder contains gas at 300 kPa at 27°C. The cylinder can withstand a maximum pressure of $1.2 \times 10^6$ Pa. If the room catches fire, what is the minimum temperature (in °C) at which the cylinder will burst?

  • 927°C✔️
  • 1127°C
  • 827°C
  • 1200°C
Correct Answer Explanation

Using Gay-Lussac's Law at constant volume:

$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$

$P_1 = 300\ \text{kPa} = 3 \times 10^5\ \text{Pa},\ T_1 = 300\ \text{K}$

$P_2 = 1.2 \times 10^6\ \text{Pa}$

$T_2 = T_1 \times \dfrac{P_2}{P_1} = 300 \times \dfrac{1.2\times10^6}{3\times10^5} = 300 \times 4 = 1200\ \text{K}$

$T_2\ (°C) = 1200 - 273 = \mathbf{927}°\text{C}$