The pressure on a sample of an ideal gas is increased from 715 mmHg to 3.55 atm

Question

The pressure on a sample of an ideal gas is increased from 715 mmHg to 3.55 atm at constant temperature. If the initial volume of the as is 472 mL. what is the final volume of the gas? A 4.00 L sample of gas is cooled from 71 CC to a temperature at which its volume is 2.60 L. What is this new temperature? Assume no change in pressure of the gas. Number Number .0014 mL 109.2 A scuba diver releases a balloon containing 121 L of helium attached to a tray of artifacts at an underwater archaeological site (see photo below). When the balloon reaches the surface, it has expanded to a volume of 333 L. The pressure at the surface is 1.00 atm; what is the pressure at the underwater site? Pressure increases by 1.0 atm for every 10 m of depth; at what depth was the diver working? Assume the temperature remains constant pressure at underwater site Number atm depth of underwater site Number

Explanation / Answer

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

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