Question

A 125 mL bubble of hot gases at 215 ∘C and 1.74 atm escapes from an active volcano.

What is the new volume of the bubble outside the volcano where the temperature is -16 ∘C and the pressure is 0.64 atm?

Answer 1

$215^oC = 215 + 273.15 = 488.15 K$

$-16^oC = -16 + 273.15 = 257.15 K$

According to the combined gas law,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2};$

$\frac{1.74atm\times125mL}{488.15K}=\frac{0.64atm\times{V_2}}{257.15K};$

$V_2=179mL$

Answer: 179 mL