A scuba diver that ascends to the surface too quickly can experience decompression sickness, which occurs when nitrogen that dissolves in the blood under high pressure, forms bubbles as the pressure decreases during the ascent. Therefore, an understanding of the gas laws is an important part of a scuba diver's training. In fresh water, the pressure increases by 1 atm every 34 ft below the water surface a diver descends.

If a diver ascends quickly to the surface from a depth of 102 ft without exhaling, by what factor will the volume of the diver's lungs change upon arrival at the surface?

The factor that will change the volume of the diver's lungs upon reaching the surface is 4

Explanation:

Given data:

Pressure increases 1 atm = 101.325 kPa

34 ft = 10.3632 m

Depth of 102 ft = 31.0896 m

Question: What factor will the volume of the diver's lungs change upon arrival at the surface, V₂/V₁ = ?

The pressure at 31.0896 m:

The factor will the volume of the diver's lungs change upon arrival at the surface:

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

i just learnt this recently but i need to look through my notes again