Question

1. scuba diver 40 ft below the ocean surface inhales 45.0 mL of compressed air from a scuba tank at a pressure of 2.90 atm and a temperature of 8 ∘ C . What is the final pressure of the air, in atmospheres, in the lungs when the gas expands to 145.0 mL at a body temperature of 37 ∘ C , and the amount of gas does not change?

2. Consider 4.70 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 3.00 L and the temperature is increased to 31 ∘C , what is the new pressure, P2 , inside the container? Assume no change in the amount of gas inside the cylinder.

3. A cylinder, with a piston pressing down with a constant pressure, is filled with 1.90 moles of a gas (n1 ), and its volume is 50.0 L (V1 ). If 0.400 mole of gas leak out, and the pressure and temperature remain the same, what is the final volume of the gas inside the cylinder?

4.A sample of gas in a cylinder as in the example in Part A has an initial volume of 52.0 L , and you have determined that it contains 1.30 moles of gas. The next day you notice that some of the gas has leaked out. The pressure and temperature remain the same, but the volume has changed to 13.0 L . How many moles of gas (n2 ) remain in the cylinder?

Answer 1

1)

Given:

Pi = 2.90 atm

Vi = 45.0 mL

Vf = 145.0 mL

Ti = 8.0 oC

= (8.0+273) K

= 281 K

Tf = 37.0 oC

= (37.0+273) K

= 310 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(2.9 atm*45 mL)/(281.0 K) = (Pf*145.0 mL)/(310 K)

Pf = 0.993 atm

Answer: 0.993 atm

2)

Given:

Pi = 365 mmHg

Vi = 4.70 L

Vf = 3.00 L

Ti = 20.0 oC

= (20.0+273) K

= 293 K

Tf = 31.0 oC

= (31.0+273) K

= 304 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(365.0 mmHg*4.7 L)/(293.0 K) = (Pf*3.0 L)/(304.0 K)

Pf = 593 mmHg

Answer: 593 mmHg