CHM 111 Problems-Gas Laws Mailings Review View NormalNo Spacing Heading 1He 5) A

Question

CHM 111 Problems-Gas Laws Mailings Review View NormalNo Spacing Heading 1He 5) A diver has 2.2 L of air in his lungs at an air temperature of 28 C and a pressure of 0.980 atm. What is the volume of air in his lungs after he dives to a depth of 10m, while holding his breath, where the temperature is 21 C and the pressure is 1.40 atm? 6) A steel tank has a volume of 438 L and is filled with 0.885 Kh of 02. Calculate the pressure of 02 at 21 O. 7) A blimp is filled with 3950 Kg of helium at 731 mmHg and 20 C. What is the volume of the blimp under these conditions? 8) Find density (g/L) of CO2 and the number of molecules per liter at 20 C and 0.9 atm. 9) Compare the density of cO2 at 0 C and 380.0 torr with its density at STP 10) An empty 149 ml flask weighs 68 g before a sample of volatile liquid is added. The flask is then placed in a hot (85C) water bath with barometric pressure at 740 torr. The liquid vaporizes and the gas fills the flask. After cooling, flask and condensed liquid together weigh 68.697 g. What is the molar mass of the liquid 11) A chemical engineer places a mixture of noble gases consisting of 5.50 g of He, 15.0 g of Ne, and 35.0g of Kr in a piston-cylinder assembly at STP. Calculate the partial pressure of each gas. 12) Solid lithium hydroxide is used to scrub CO2 from the air in spacecraft submarines. It reacts with CO2 to produce lithium carbonate and water. What mass of lithium hydroxide is required to remove 215 L of CO2 at 23 C and 0.942 atm.2 2LiOH(s) +CO2(g)-- Li2CO3(s) +H2O(l) Focus

Explanation / Answer

Ans 5 :

Using the ideal gas law , pV = nRT

We get the relation as :

p1V1 T2 = p2V2 T1

T1 = 28 C = 301.15 K

T2 = 21 C = 294.15 K

So putting all the values we get :

0.980 x 2.2 x 294.15 = 1.40 x V2 x 301.15

V2 = 1.5 L

So the volume will be 1.5 L.

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