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Gas Laws Fact Sheet

| Property | Value |
|---------------------------------|----------------------------------|
| Ideal gas law | [tex]\( PV = nRT \)[/tex] |
| Ideal gas constant | [tex]\( R = 8.314 \frac{L \cdot kPa}{mol \cdot K} \)[/tex] or [tex]\( R = 0.0821 \frac{L \cdot atm}{mol \cdot K} \)[/tex] |
| Standard atmospheric pressure | [tex]\( 1 \, atm = 101.3 \, kPa \)[/tex] |
| Celsius to Kelvin conversion | [tex]\( K = ^\circ C + 273.15 \)[/tex] |

Question:
A scuba diver's air tank contains oxygen, helium, and nitrogen at a total pressure of 205 atmospheres. The partial pressure of nitrogen is 143 atmospheres, and the partial pressure of helium is 41 atmospheres. What is the partial pressure of oxygen in the tank?

A. 21 atm
B. 103 atm
C. 307 atm
D. 389 atm


Sagot :

To determine the partial pressure of oxygen in the scuba diver's air tank, we will use Dalton's Law of Partial Pressures. According to this law, the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases. Mathematically, it can be expressed as:

[tex]\[ P_{\text{total}} = P_{\text{nitrogen}} + P_{\text{helium}} + P_{\text{oxygen}} \][/tex]

Let’s break this down step-by-step using the given values:

1. Total pressure, [tex]\( P_{\text{total}} \)[/tex], is 205 atmospheres.
2. Partial pressure of nitrogen, [tex]\( P_{\text{nitrogen}} \)[/tex], is 143 atmospheres.
3. Partial pressure of helium, [tex]\( P_{\text{helium}} \)[/tex], is 41 atmospheres.

We need to find the partial pressure of oxygen, [tex]\( P_{\text{oxygen}} \)[/tex]. Using the formula from Dalton's Law:

[tex]\[ P_{\text{total}} = P_{\text{nitrogen}} + P_{\text{helium}} + P_{\text{oxygen}} \][/tex]

Substitute the known values into the equation:

[tex]\[ 205 = 143 + 41 + P_{\text{oxygen}} \][/tex]

We can isolate [tex]\( P_{\text{oxygen}} \)[/tex] by rearranging the equation:

[tex]\[ P_{\text{oxygen}} = 205 - 143 - 41 \][/tex]

Subtract the partial pressures of nitrogen and helium from the total pressure:

[tex]\[ P_{\text{oxygen}} = 205 - 143 - 41 \][/tex]
[tex]\[ P_{\text{oxygen}} = 21 \][/tex]

Thus, the partial pressure of oxygen in the tank is [tex]\( 21 \)[/tex] atmospheres, which matches option A.

Therefore, the correct answer is:

A. [tex]\( 21 \)[/tex] atm