A125 ml bubble of hot gases at 213 ∘c and 1.77 atm is emitted from an active volcano. part a what is the final temperature, in degrees celsius, of the gas in the bubble outside the volcano if the final volume of the bubble is 136 ml and the pressure is 0.810 atm , if the amount of gas does not change?

T₂  = 242 K

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 125 mL

Initial pressure = 1.77 atm

Initial temperature = 213 °C (213 +273 = 486 K)

Final volume = 136 mL

Final pressure = 0.810 atm

Final temperature = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

T₂ = P₂V₂T₁ / P₁V₁

T₂ = 0.810 atm × 136 mL × 486 K / 1.77 atm× 125 mL

T₂  =  53537.76 atm .mL.  K  / 221.25 atm . mL

T₂  = 242 K

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