1/ a. A scuba diver 60 ft below the ocean surface inhales 70.0 mL of compressed

Question

1/ a. A scuba diver 60 ft below the ocean surface inhales 70.0 mL of compressed air from a scuba tank at an initial pressure of 3.20 atm and temperature of 5 C. What is the final pressure of air, in atmo-spheres, in the lungs when the gas expands to 170.0 mL at a body temperature of 37 C, and the amount of gas remains constant?

b. 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 48.0 L (V1). If 0.800 mole of gas leak out, and the pressure and temperature remain the same, what is the final volume of the gas inside the cylinder?

c.A sample of gas in a cylinder as in the example in Part A has an initial volume of 44.0 L , and you have determined that it contains 1.20 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 11.0 L . How many moles of gas (n2) remain in the cylinder?

Explanation / Answer

1.Given P1= 3.2 atm, V1= 70ml= (70/1000) L, T1= 5 deg.c= 5+273= 278 K,

Given V2= 170ml = (170/1000) L, T2= 37deg.c= 37+273=310K,

From gas law, P1V1/T1= P2V2/T2, P2= P1V1T2/(T1V2) = 3.2*(70/1000)*310/(278*170/1000)= 2.5 atm

2.

2. since pressure and temperature remains the same

PV= nRT, P/T= nR/V = constant

Or n1R/V1= n2R/V2, n= moles of gas at initial condition= 1.9 moles. n2= mole of gas at final condition= 1.9-0.8=1.1 moles and V1=48 L

Hence 1.9/48= 1.1/V2

V2= 1.1*48/1.9= 27.78 L

3. let n2 -mole of gas remaining

hence n1/V1= n2/V2

1.2/44= n2/11

n2= 1.2*11/44= 0.3 moles remain

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