1. At an underwater depth of 255 ft , the pressure is 8.36 atm . What should the
ID: 636249 • Letter: 1
Question
1. At an underwater depth of 255 ft , the pressure is 8.36 atm .
What should the mole percent of oxygen be in the diving gas for the partial pressure of oxygen in the mixture to be 0.21 atm, the same as in air at 1 atm?
2. A gaseous mixture of O2 and N2 contains 37.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 845 mmHg ?
Express you answer numerically in millimeters of mercury.
3.A certain amount of chlorine gas was placed inside a cylinder with a movable piston at one end. The initial volume was 3.00 L and the initial pressure of chlorine was 1.55 atm . The piston was pushed down to change the volume to 1.00 L. Calculate the final pressure of the gas if the temperature and number of moles of chlorine remain constant.
4. A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 3.90 L ? (The temperature was held constant.)
5. 15.0 moles of gas are in a 7.00 L tank at 24.2 ?C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2?atm/mol2 and b=0.0430 L/mol.
Express your answer with the appropriate units.
Explanation / Answer
1. Mole % of oxygen in the mixture is proportional to its partial pressure
= P[O2]/P[T] *100
=0.21/8.36 *100
=2.5%
2. For this we have to find out first the partial molar ratio of oxygen and nitrogen
which is calculated as
moles of N2 =37.8/28=1.35
moles of O2 =62.2/32 =1.94
TOATAL NUMBER OF MOLES=3.29
partial pressure of O2 =1.94/3.29 =0.589
partial pressure of O2 respect to mercury
0.589*845 =497.7mm og Hg
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