Chapter 8.5, Problem 8.38QAP

Interpretation Introduction

Interpretation:

The final pressure of air in the lungs of a scuba diver who is 60ft below the ocean surface should be calculated if the final volume of air is 150.0 ml at a temperature of37°C. Given that the air inhaled by the diver has an initial volume of 50.0 ml, pressure of 3.00 atm and at a temperature of8°C

Concept introduction:

► As per Charles law, volume is directly related to temperature:

$\begin{array}{l}\text{V α T i}\text{.e}\text{. V = k}\times \text{T}\\ \text{where 'k' is the proportionality constant}\\ (\text{or})\frac{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ = }\frac{{\text{V}}_2}{{\text{T}}_2}-----(1)\\ {\text{where V}}_{\text{1}}{\text{ and V}}_{\text{2}}\text{ are the initial and final volume}\\ {\text{T}}_{\text{1}}{\text{ and T}}_{\text{2}}\text{ are the initial and final temperature}\end{array}$

► As per Boyles law, pressure is inversely related to volume:

$\begin{array}{l}\text{P α }\frac{\text{1}}{\text{V}}\text{ i}\text{.e}\text{. P = k}\times \frac{\text{1}}{\text{T}}\\ \text{where k is a proportionality constant}\\ {\text{(or) P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = P}}_{\text{2}}{\text{V}}_{\text{2}}\text{ -----(2)}\end{array}$

► As per Gay-Lussac’s law, pressure is directly related to temperature:

$\begin{array}{l}\text{P α T i}\text{.e}\text{. P = k}\times \text{T}\\ \text{where 'k' is the proportionality constant}\\ (\text{or})\frac{{\text{P}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ = }\frac{{\text{P}}_{\text{2}}}{{\text{T}}_{\text{2}}}-----(3)\\ {\text{where P}}_{\text{1}}{\text{ and P}}_{\text{2}}\text{ are the initial and final pressures}\\ {\text{T}}_{\text{1}}{\text{ and T}}_{\text{2}}\text{ are the initial and final temperature}\end{array}$

► Equations (1), (2) and (3) give the combined gas law according to which:

$\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{ = }\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}\text{--------(4)}$

► Volume conversion:

1 ml = 0.001L (or) 1L = 1000 ml

Temperature conversion:

K = °C + 273 (or) °C = K - 273

Pressure conversion:

1 atm = 760 torr = 760 mmHg

1 torr = 1 mmHg

Given:

Initial volume V₁ = 50.0 mL

Initial pressure P₁ = 3.00 atm

Initial temperature T₁ = 8°C = (8 + 273) K = 281 K

Final temperature T₂ = 37°C = (37 + 273) K = 310 K

Final volume = V₂ = 150.0 mL

Final pressure = P₂