Question

An average human adult takes about 15breaths per min and exhales 23ml of co2 with each breath. If your trapped in a cave with pressure .963atm and 11C and your only source of oxygen is a breathing device containing 1kg of KO2 , how long do you have to live?

Answer 1

1

Answer to the Question#47310-Chemistry, Other

Question:

An average human adult takes about 15 breaths per min and exhales 23ml of CO₂ with each breath. If your are trapped in a cave with pressure 0.963 atm and 11°C and your only source of oxygen is a breathing device containing 1 kg of KO₂, how long do you have to live?

Solution:

KO₂ (potassium superoxide) can be used for rebreathers (CO₂-scrubbers). The KO₂-rebreather units work by converting the exhaled CO₂ to O₂ via the following chemical reaction:

$$4 \mathrm{KO}_2 + 2 \mathrm{CO}_2 = 2 \mathrm{K}_2 \mathrm{CO}_3 + 3 \mathrm{O}_2$$

Notice the potassium superoxide consumes CO₂, a product of respiration, and produces O₂, which can be breathed.

To solve our problem, we can describe breathing by simplified chemical equation:

$$\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6 \mathrm{O}_2 = 6 \mathrm{CO}_2 + 6 \mathrm{H}_2\mathrm{O}$$

1. Now calculate moles of CO₂ exhales per 1 breath using the ideal gas equation:

$$\mathrm{PV} = nRT,$$

Convert all values to standard form:

$$0.963 \mathrm{atm} = (101325 \mathrm{Pa/atm}) \times 0.963 \mathrm{atm} = 97576 \mathrm{Pa}, 23 \mathrm{ml} = 2.3 \times 10^{-5} \mathrm{m}^3, 11^{\circ} \mathrm{C} = 284 \mathrm{K},$$ $$R = 8.314 \mathrm{Pa} \times \mathrm{m}^3/\mathrm{mol} \times \mathrm{K}$$

Per 1 breath: $n = \mathrm{PV} / \mathrm{RT} = 97576 \times 2.3 \times 10^{-5} / 8.314 \times 284 = 9.5 \times 10^{-4} \mathrm{mol}$

2. How many moles of O₂ is needed to 1 breath?

According to simplified chemical equation of breathing 1 mole of O₂ are equals to 1 mole of CO₂. Thus $9.5 \times 10^{-4} \mathrm{mol}$ of O₂ is needed to 1 breath.

3. How many moles of O₂ can be obtained from 1 kg (1000 g) of KO₂?

KO₂ reacts with CO₂ according to the equation:

$$4 \mathrm{KO}_2 + 2 \mathrm{CO}_2 = 2 \mathrm{K}_2 \mathrm{CO}_3 + 3 \mathrm{O}_2$$

As we can see, 1 mole of O₂ is equivalent to 4/3 moles of KO₂

$$M(\mathrm{KO}_2) = 39 + 16 \times 2 = 71 \ \mathrm{g/mol},$$ $$4/3 \ \text{moles of } \mathrm{KO}_2 = 71 \times 4/3 = 94.7 \ \mathrm{g/mol} \ \mathrm{O}_2$$ $$1000 \ \mathrm{g} / (94.7 \ \mathrm{g/mol} \ \mathrm{O}_2) = 10.6 \ \mathrm{mol} \ \mathrm{O}_2$$

4. How many breaths you can do with 10.6 mol O₂?

$$10.6 \ \mathrm{mol} / 9.5 \times 10^{-4} \ \mathrm{mol} = 11157 \ \text{breaths}$$

5. How long it will be possible to breathe?

$$11157 \ \text{breaths} / (15 \ \text{breaths/min}) = 743 \ \text{min}$$

Answer: You can to live about 743 minutes.