In: Chemistry
Part A) For the first we use,
P1V1 = P2V2
Given are,
P1 = 715 mmHg = 0.941 atm
V1 = 472 mL
P2 = 3.55 atm
V2 = unknown
Thus,
0.941 x 472 = 3.55 x V2
V2 = 125.11 mL
Thus 125.11 mL is the final volume of the gas.
Part B) For the second we use,
V1/T1 = V2/T2
We have,
V1 = 4.00 L
T1 = 71 oC
V2 = 2.60 L
T2 = unknown
Therefore,
4/71 = 2.6/T2
T2 = 46.15 oC
Thus the new temperature is 46.15 oC
Part C) We will use
P1V1 = P2V2
Given are,
P1 = unknown
V1 = 121 L
P2 = 1.0 atm
V2 = 333 L
thus,
121 x P1 = 1 x 333
P1 = 2.75 atm
therefore, the pressure at the underwater level is 2.75 atm
Now,
P = 1 + d/33
feed P value at underwater level,
2.75 = 1 + d/33
d = 57.75 ft = 17.60 m
thus, the dept of diver is 17.60 m