Question

6) A whale is able to dive to depths of 1000 meters, where the pressure is 100 atm and the temperature is 20^oF. If the whale takes a breath and fills its lungs with a volume of 400 L at the surface, where the temperature is 65^oF, and the pressure is 1 atm, What is the volume of its lungs at this depth?

7) I decide to blow up a balloon. The volume initially was 21.2-L and it changes to 39.5 L. The balloon initially weighed 1.35 g and was filled with oxygen gas. I added hydrogen gas to reach the new volume. How many grams of hydrogen gas musthave I added? Assume constant temperature and a pressure of 758 mmHg.

8) Calculate the constant temperature in 7) ? Does this make sense given the amount of gas and the volume of the balloons?

Answer 1

6) in depth of the sea

P2 = 100 atm , V2 = ? L, T2 = 20 F

at the surface

P1 = 1 atm V1 = 400 L, T1 = 65 F

from ideal gas equation

P1V1/T1 = P2V2/T2

V2 = (1*400/65)*(20/100) = 1.23 L

volume of its lungs at this depth = 1.23 L

7)

initia volume V1 = 21.2 L

final volume V2   = 39.5 L

at constant T,P

No of moles of O2 filled in ballon = 1.35/32 = 0.0421875 mole

from ideal gas equation :

V1/n1 = V2/n2

21.2/0.0421875 = 39.5/n2

n2 = 0.0786 mole

No of moles of H2 added = 0.0786-0.0421875

            = 0.0364125 mole

mass of H2 = 0.0364125 * 2 = 0.072825 grams

8) T = PV/nR

n = total no of moles = 0.0786 mole

P = 758 mmhg = 0.99737 Atm

V = final volume   = 39.5 L

R = gas constant = 0.0821 l.atm.k-1.mol-1

T = (0.99737*39.5)/(0.0786*0.0821)

= 6105.03 K

= 5832.03 C

it doesnt make any sense.