A100.0 ml bubble of hot gases at 225 °c and 1.80 atm escapes from an
active volcano, what is the new volume of the bubble outside the volcano
where the temperature is -25 °c and the pressure is 0.80 atm?

The new or final volume is 112.1 mL.

Explanation:

Given data:

Initial temperature = 225 °C  (225 °C + 273.15 = 498.15 K)

Initial pressure = 1.80 atm

Initial volume = 100.0 mL

Final temperature = -25 °C  (-25 °C + 273.15 = 248.15 K)

Final pressure = 0.8 atm

Final volume = ?

Formula

P₁V₁ / T₁   = P₂V₂ /T₂

Solution:

V₂ =  P₁V₁. T₂  / T₁ .  P₂

V₂ =  1.80 atm. 100 mL . 248.15 K / 498.15 K .  0.8 atm

V₂ =  44667 mL /398.52

V₂ = 112.1 mL

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey