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The average lung capacity of a human is 6.0 L. How many moles of air are in your lungs in the following situations? Assume air behaves ideally and has an average molar mass of [tex]$29.0 g / mol$[/tex].

a. At sea level [tex](T=298 \, \text{K}, P =1.00 \, \text{atm})[/tex].
Moles = [tex]$\square$[/tex] mol air

b. 10 m below water [tex](T=298 \, \text{K}, P =1.98 \, \text{atm})[/tex].
Moles = [tex]$\square$[/tex] mol air

c. At the top of Mount Everest [tex](T=203 \, \text{K}, P =0.297 \, \text{atm})[/tex].
Moles = [tex]$\square$[/tex] mol air

Sagot :

GinnyAnsweridn
To solve for the number of moles in each situation, we will use the ideal gas law equation:

[tex]\[ PV = nRT \][/tex]

where:
- [tex]\( P \)[/tex] is the pressure,
- [tex]\( V \)[/tex] is the volume,
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the ideal gas constant (0.0821 L·atm/(K·mol)),
- [tex]\( T \)[/tex] is the temperature in Kelvin.

We can rearrange the equation to solve for [tex]\( n \)[/tex]:

[tex]\[ n = \frac{PV}{RT} \][/tex]

### a. At sea level [tex]\((T = 298 K, P = 1.00 atm)\)[/tex]:

1. Variables:
- [tex]\( P = 1.00 \)[/tex] atm
- [tex]\( V = 6.0 \)[/tex] L
- [tex]\( T = 298 \)[/tex] K
- [tex]\( R = 0.0821 \)[/tex] L·atm/(K·mol)

2. Calculation:
[tex]\[ n = \frac{(1.00 \text{ atm}) \times (6.0 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)}) \times (298 \text{ K})} \][/tex]
[tex]\[ n \approx 0.24524029461534058 \text{ mol} \][/tex]

Thus, the number of moles of air in your lungs at sea level is [tex]\( \approx 0.245 \)[/tex] mol air.

### b. 10 m below water [tex]\((T = 298 K, P = 1.98 atm)\)[/tex]:

1. Variables:
- [tex]\( P = 1.98 \)[/tex] atm
- [tex]\( V = 6.0 \)[/tex] L
- [tex]\( T = 298 \)[/tex] K
- [tex]\( R = 0.0821 \)[/tex] L·atm/(K·mol)

2. Calculation:
[tex]\[ n = \frac{(1.98 \text{ atm}) \times (6.0 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)}) \times (298 \text{ K})} \][/tex]
[tex]\[ n \approx 0.4855757833383743 \text{ mol} \][/tex]

Thus, the number of moles of air in your lungs 10 meters below water is [tex]\( \approx 0.4856 \)[/tex] mol air.

### c. At the top of Mount Everest [tex]\((T = 203 K, P = 0.297 atm)\)[/tex]:

1. Variables:
- [tex]\( P = 0.297 \)[/tex] atm
- [tex]\( V = 6.0 \)[/tex] L
- [tex]\( T = 203 \)[/tex] K
- [tex]\( R = 0.0821 \)[/tex] L·atm/(K·mol)

2. Calculation:
[tex]\[ n = \frac{(0.297 \text{ atm}) \times (6.0 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)}) \times (203 \text{ K})} \][/tex]
[tex]\[ n \approx 0.10692235229175043 \text{ mol} \][/tex]

Thus, the number of moles of air in your lungs at the top of Mount Everest is [tex]\( \approx 0.1069 \)[/tex] mol air.
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