A small research submarine with a volume of 1.2x10 ^5 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?

Dec 7, 2014

The answer is $573 L$.

The best way to approach this problem is by using the combined gas law

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$, which relates temperature, pressure, and volume, assuming the number of moles remains constant. Knowing that

${P}_{1} = 1 a t m$, ${T}_{1} = 273.15 + 150 = 423.15 K$, ${V}_{1} = 1.2 \cdot {10}^{5} L$, ${P}_{2} = 150 a t m$, and ${T}_{2} = 303.15 K$, we get

${V}_{2} = {P}_{1} / {P}_{2} \cdot {T}_{2} / {T}_{1} \cdot {V}_{1} = \frac{1}{150} \cdot \frac{303.15}{423.15} \cdot 1.2 \cdot {10}^{5} = 573 L$

Notice how both a significant increase in pressure and an important drop in temperature contributed to the massive drop in volume the submarine underwent.