Question

a certain amount of gas at 23.0 °c and at a pressure of 0.950 atm is contained in a glass vessel. suppose that the vessel can withstand a pressure of 2.30 atm. how high can you raise the temperature of the gas without bursting the vessel? in other words, at what temperature will the glass vessel shatter? round your answer to 3 significant digits.

Explanation:

Step1: Convert initial temp to Kelvin

$T_1 = 23.0^\circ\text{C} + 273.15 = 296.15\ \text{K}$

Step2: Apply Gay-Lussac's Law

$\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies T_2 = T_1 \times \frac{P_2}{P_1}$
$T_2 = 296.15\ \text{K} \times \frac{2.30\ \text{atm}}{0.950\ \text{atm}} \approx 712.8158\ \text{K}$

Step3: Convert back to Celsius

$T_2^\circ\text{C} = 712.8158 - 273.15 \approx 439.6658^\circ\text{C}$

Step4: Round to 3 sig figs

$440^\circ\text{C}$ (3 significant digits)

Answer:

440