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?

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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?

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Answer 1 · 2014-12-07T23:02:55

The answer is $573 L$ $573L$ .

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

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1 = (P_2V_2)/T_2$ , which relates temperature, pressure, and volume, assuming the number of moles remains constant. Knowing that

$P_{1} = 1 a t m$ $P_1 = 1atm$ , $T_{1} = 273.15 + 150 = 423.15 K$ $T_1 = 273.15 + 150 = 423.15K$ , $V_{1} = 1.2 \cdot 10^{5} L$ $V_1 = 1.2* 10^5 L$ , $P_{2} = 150 a t m$ $P_2 = 150 atm$ , and $T_{2} = 303.15 K$ $T_2 = 303.15K$ , we get

$V_{2} = \frac{P_{1}}{P_{2}} \cdot \frac{T_{2}}{T_{1}} \cdot V_{1} = \frac{1}{150} \cdot \frac{303.15}{423.15} \cdot 1.2 \cdot 10^{5} = 573 L$ $V_2 = P_1/P_2 * T_2/T_1 * V_1 = 1/150 * (303.15)/(423.15) * 1.2 * 10^5 = 573L$

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

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