Q. 8.46 - An Introduction to General, Organic, and Biological Chemistry

Question

A scuba diver $60 ft$ below the ocean surface inhales $50.0 mL$of compressed air from a ccuba tank at a pressure of $3.00 atm$ and a temperature of $8°C$. What is the final pressure of air, in atmospheres, in the lungs when the gas expands to $150.0 mL$ at a body temperature of $37 °C$, if the amount of gas does not change?

Step-by-Step Solution

Answer

The final pressure of the gas is $1.1 atm$.

Step 1: Given Information

We need to find the final pressure of gas.

Step 2: Simplify

The combined gas law is:

$\frac{P_1\times V_1}{T_1}=\frac{P_2\times V_2}{T_2}$

Consider:

initial pressure $P_1=3 atm$

initial temperature $T_1=8°C=\left(8+273\right)K=281 K$

initial volume $V_1=50 mL$

final temperature $T_2=37°C=\left(37+273\right)K=310 K$

final volume $V_2=150 mL$

Now, calculating the final pressure $P_2$ is:

$$\begin{gathered} P_2=\left(\frac{\left(P_1\times V_1\right)}{T_1}\right)\times \left(\frac{V_2}{T_2}\right) \\ {P}_2=\left(\frac{\left(3 atm \times 50 mL\right)}{281 K}\right)\times \left(\frac{310 K}{150 mL}\right) \\ {P}_2=1.1 atm \end{gathered}$$