Question

A diver exhales a bubble with a volume of 250 mL at a pressure of 2.4 atm and a temperature of 15 °C. What is the volume of the bubble when it reaches the surface where the pressure is 1.0 atm and the temperature is 27 °C? A. 630mLB. 110 mLC. 580 mLD. 1100 mLE. 100 mL

Answer 1

`\large \boxed{\text{A. 620mL}}`

Step-by-step explanation:

We can use the Combined Gas Laws to solve this problem

Data

p₁ = 2.4 atm; p₂ = 1.0 atm

V₁ = 250 mL; V₂ = ?

T₁ = 15°C; T₂ = 27 °C

Calculations

(a) Convert the temperatures to kelvins

T₁ = (15 + 273.15) K = 288.15 K

T₂ = (27 + 273.15) K = 300.15 K

(b) Calculate the new volume

`\begin{array}{rcl}(p_(1)V_(1) )/(T_(1)) & = & (p_(2)V_(2))/(T_(2))\\\\\frac{\text{2.4 atm $*$ 250 mL}}{\text{288.15 K}} & = & \frac{\text{1.0 atm} * V_(2)}{\text{300.15 K}}\\\\\text{2.08 mL} & = & (V_(2))/(300.15)\\\\V_(2) & = & \textbf{620 mL}\\\end{array}\\\text{The volume of the bubble when it reaches the surface is $\large \boxed{\textbf{620 mL}}$}`

Answer 2

A. 630mL

Step-by-step explanation:

The combined gas law says:

P₁V₁/T₁ = P₂V₂/T₂

Where P₁ is 2,4 atm; V₁ is 250mL; T₁ is 15+273,15 = 288,15K; P₂ is 1,0 atm; V₂ is the final volume; T₂ is 27°C; 27+273,15 = 300,15K.

Thus:

2,4atm×250mL/288,15K = 1,0atm×V₂/300,15K

V₂ = 625mL ≈ A. 630mL

I hope it helps!