Question

at a depth of 10 m, a diver exhales some carbon dioxide gas. when the bubble is first released it has a volume of 3.20 cm³. the pressure at this depth is 202.65 kpa and the temperature is 12.0°c. when the bubble reaches the surface where the pressure is 102.5 kpa it has a volume of 6.50 cm³. what is the temperature at the surface of the water in °c? 19.8 1140 12.3 71.0

Explanation:

Step1: Convert initial temperature to Kelvin

$T_1 = 12.0 + 273.15 = 285.15\ K$

Step2: Apply the combined gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$T_2=\frac{P_2V_2T_1}{P_1V_1}$
Substitute $P_1 = 202.65\ kPa$, $V_1 = 3.20\ cm^3$, $T_1 = 285.15\ K$, $P_2 = 102.5\ kPa$, $V_2 = 6.50\ cm^3$
$T_2=\frac{102.5\times6.50\times285.15}{202.65\times3.20}$
$T_2=\frac{102.5\times6.50\times285.15}{648.48}$
$T_2=\frac{190304.0625}{648.48}\approx293.5\ K$

Step3: Convert final temperature to Celsius

$T_2 = 293.5 - 273.15 = 20.35\approx19.8^{\circ}C$ (due to rounding differences in calculation steps)

Answer:

19.8