A small research submarine with a volume of 1.2 x 105 L has an internal pressure of 1.0 atm and an internal temperature of 150 C. If the submarine descends to a depth where the pressure is 150 atm and the temperature is 30 C, what will the volume of the gas inside be if the hull of the submarine breaks?

V₂ = 570 L

Explanation:

Given data:

Initial volume = 1.2 ×10⁵ L

Initial pressure = 1 atm

Initial temperature = 150 °C (150 +273 = 423 K)

Final temperature = 30°C (30+273 = 303 K)

Final volume = ?

Final pressure = 150 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1.0 atm × 1.2 ×10⁵ L × 303 K / 423 K × 150 atm

V₂ = 363.6×10⁵ L atm .L. K / 63450 K.atm

V₂ = 0.0057×10⁵ L

V₂ = 570 L

h let me think