Question

The enclosed cabin of a submarine has a volume of 2.4 × 105 liters, a temperature of 312 K, and a pressure of 116 kPa. As people in the cabin breathe, carbon dioxide gas, CO2(g), can build up to unsafe levels. Air in the cabin becomes unsafe to breathe when the mass of CO2(g) in this cabin exceeds 2156 grams. Show a numerical setup for calculating the pressure in the submarine cabin if the cabin temperature changes to 293 K.

Answer 1

To solve this problem we will use the ideal gas equations. First we will use the equation to determine the number of moles with the given pressure. As the matter is conserved, the gas pressure is subsequently calculated in state two. Let's start defining our variables

Volume, $V=2.4*10^5L$

Pressure, $P = 116kPa$

Temperature, $T=312K$

By Ideal gas law

$PV = nRT$

$n = \frac{PV}{RT}$

So, moles of gas, n,

$n = \frac{(116)(2.4*10^5)}{(8.314)(312)}$

$n = 1.073*10^4mol$

Now, when the temperature is changed to 293K

$PV=nRT$

$P = \frac{nRT}{V}$

$P = \frac{(1.073*10^4)(8.314)(293)}{2.4*10^5}$

$P = 108.9kPa$

The pressure in the submarine when the temperature is changed to 293 is 108.9kPa

Answer 2

In a 2.4 × 10⁵ L enclosed cabin, the pressure is 116 kPa at 312 K. If the temperature changes to 293 K, the pressure changes to 109 kPa.

The enclosed cabin of a submarine has a volume of 2.4 × 10⁵ liters, at a temperature of 312 K (T₁), and a pressure of 116 kPa (P₁).

If we decrease the temperature to 293 K (T₂), at constant volume, we can calculate the new pressure (P₂) using Charles' law.

$\frac{P_1}{T_1} = \frac{P_2}{T_2} \\\\P_2 = \frac{P_1 \times T_2 }{T_1} = \frac{116kPa \times 293K }{312K} = 109kPa$

In a 2.4 × 10⁵ L enclosed cabin, the pressure is 116 kPa at 312 K. If the temperature changes to 293 K, the pressure changes to 109 kPa.

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