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

the product of the pressure and volume is also constant

pv=nrt

explanation: