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My name is Ann [436]
3 years ago
8

We know that standard pressure is one atmosphere, or 760 millimeters of mercury. This pressure results from the weight of gas mo

lecules in the atmosphere. As a diver enters the water, he is subject to both water pressure and air pressure. Because water is much denser than air, the pressure increases rapidly as the diver descends. At the depth of 34 feet in fresh water, the diver is experiencing 2 atmospheres of pressure (one from air pressure and one from the 34 feet of water). For every additional 34 feet the diver descends he will be under an additional atmosphere of pressure. Since water pressure is proportional to depth, how many atmospheres of pressure would a diver experience at 102 feet? Why wouldn't this pressure squash the diver? Answering this second question may be easier if you think of the reason a person on land is not squashed by one atmosphere of pressure. Explain your answer in detail.
Chemistry
1 answer:
Vladimir79 [104]3 years ago
7 0
Q1)
When the diver descends 34 feet;
total pressure on the diver - air pressure+ pressure from 34 feet of water 
air pressure - 1 atm
water pressure - 1 atm
therefore total pressure at 34 feet deep - 2 atm
Since water pressure increases with depth, according to the following equation;
P = hpg
h - height of water column above the point of pressure 
p - density of water
g - gravitational acceleration
P - pressure
as p and g are constant P is proportional to height, therefore when the height increases, pressure too increases proportionally
new height is 102 feet
that's 34*3 = 102 feet
so for every 34 feet, pressure increases by 1 atm
therefore at 102 feet - 1 atm times 3 = 3 atm
therefore total pressure at the new point = 1 atm (air pressure) + 3 atm (water pressure)  
total pressure = 4 atm

Q2)the reason why a person on land is not affected by the pressure exerted on him by the atmosphere around him is as the persons body too exerts an almost equal amount of pressure from inside. Therefore the pressure acting on him and pressure exerted by him is equalised so he wont feel the effect of atmospheric pressure.
Similarly when the person dives into water since the body also contains water inside, the pressure exerted on him by water is equal to the internal pressure coming from him. Therefore he won't feel a huge effect by the water pressure.
However the air spaces inside the person will have the pressure it had at the surface of water, but when the person has dived inside the water pressure is much higher therefore there's a pressure difference between that of the water pressure and pressure inside the air spaces. This may cause a squeeze where the air spaces tend to compress. To avoid this condition, the air spaces should be equalised.
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As 4 moles of gaseous reactants are changing to 2 moles of gaseous products,  the randomness is decreasing and the entropy is negative

2. 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

As 9 moles of gaseous reactants are changing to 10 moles of gaseous products,  the randomness is increasing and the entropy is positive.

3. NH_4I(s)\rightarrow NH_3(g)+HI(g)

As 1 mole of solid reactants is changing to 2 moles of gaseous products, the randomness is increasing and the entropy is positive.

4. 2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)

As 4 moles of gaseous reactants is changing to 5 moles of gaseous products, the randomness is increasing and the entropy is positive

5. 2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(l)

As 4 moles of gaseous reactants is changing to 1 moles of gaseous products, the randomness is decreasing and the entropy is negative.

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Consider a 100-gram sample of radioactive cobalt-60.
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It will take 5.2 years to decay.

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what do you mean by radioactive decay ?

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Learn more about decay here:-

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