Question

1.18

part A) If the partial pressure of oxygen drops below 110 mmHg , passengers become drowsy. If this happens, oxygen masks are released. What is the total cabin pressure at which oxygen masks are dropped?

part B) A sample of gas in a cylinder as in the example in Part A has an initial volume of 42.0 L , and you have determined that it contains 1.70 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 10.5 L . How many moles of gas (n2) remain in the cylinder?

1 21 a) A very flexible helium-filled balloon is released from the ground into the air at 20. ∘C. The initial volume of the balloon is 5.00 L, and the pressure is 760. mmHg. The balloon ascends to an altitude of 20 km, where the pressure is 76.0 mmHg and the temperature is −50. ∘C. What is the new volume, V2, of the balloon in liters, assuming it doesn't break or leak?

b) Consider 5.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.40 L and the temperature is increased to 34 ∘C , what is the new pressure, P2, inside the container? Assume no change in the amount of gas inside the cylinder.

Answer 1

1.18

A.

we know that mole fraction of oxygen in air = 0.21

partial pressure of oxygen(P_O2) = P_T x Y_O2 = P_T x 0.21

P_T = total pressure

Y_O2 = mole fraction of oxygen

P_O2 = 110mmHg = P_T x 0.21

P_T = 523.81 mmHg Ans (a)

(b)

for any gas

P₁V₁ /n₁T₁ = P₂V₂/n₂T₂

where P is pressure

V = volume

n = moles of gas

T = temperature

subscript 1 and 2 represents conditions at two situation

given P₁ = P₂ same pressure

T₁ = T₂ same temperature

P₁V₁ /n₁T₁ = P₁V₂/n₂T₁

n₂ = V₂/V₁ x n₁ = 10.5 L / 42.0 L x 1.7 mol = 0.425 mol

1.21

when no. of moles is same that is n₁ = n₂ when there is no leakage of gas

according to ideal gas equation

P₁V₁ /T₁ = P₂V₂/T₂

20 ^oC = 273 + 20 = 293 K = T₁

-50 ^oC = 273 - 50 = 223 K = T₂

P₁ = 760 mmHg

P₂ = 76 mmHg

V₁ = 5.00L

V₂ = P₁V₁ T₂/T₁ P₂ = 760 mmHg x 5.00 L x 223K / (293 K x 76 mmHg) = 38.05 L

similarly as above

T₁ = 293 K

T₂ = 34 + 273 = 307K

V₁ = 5.00 L

V₂ = 2.40 L

P₁ = 365 mmHg

P₂  = P₁V₁ T₂/T₁ V₂ = 365 mmHg x 5.00 L x 307 K / (293 K x 2.40 L) = 796.75 mmHg

Ans = 796.75 mmHg