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  4. How are partial pressures of gases calculated?
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How are partial pressures of gases calculated?

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Offline siewwen168 (OP)

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How are partial pressures of gases calculated?
« on: 22/05/2005 05:08:18 »
1.) A container contains 60cm3 of CO2 and 40cm3 of O2 at room temperature and pressure conditions. If 200cm3 of H2 at room temperature and pressure is now injected into the container, which of the following would happen?[:)]
(a) The partial pressure of O2 will decrease
(b) The total pressure of the gases in the container will increase
(c) The partial pressure of CO2 and O2 remain constant
The answer given is (b) & (c),why?how to count?Please show the calculations.[:D]

2.) When a few crystals of ice are added to supercooled water at -5 degree Celcius,which of the following would happen?[:)]
(a) More ice is precipitated
(b) The vapour pressure increases
(c) The temperature of the water increases
The answer given is (a),(b)&(c),why?Please provide me some explanation.[:p]
   
« Last Edit: 20/06/2008 15:51:16 by chris »
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Offline Atomic-S

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How are partial pressures of gases calculated?
« Reply #1 on: 11/07/2008 02:52:28 »
 1: (a) would be essentially false because the amount of free oxygen is not affected by the addition of an additional gas if they do not react, and nothing in the problem mentions ignition. So, pursuant to the law of partial pressures in ideal gasses, the oxygen retains its former partial pressure.  {b) is correct because the addition of the hydrogen adds its partial pressure to the partial pressures already present, thereby raising the total pressure of the system. (C) is correct also for substantially the same reason: no chemical reaction is mentioned has taking place, so the amount of the original gasses remains as before.

2: (a) is correct because the tossing of an ice crystal into the supercooled substance provides a point of nucleation, on which the water, already below its normal freezing point, can begin crystallizing.  (b) I would be unsure about. (c) probably can be explained as the release of energy as the water assumes its perferred state, which typically would be a state of lower energy. The excess energy then would appear as heat. However, in saying this, it is necessary to be somewhat tentative, because in reactions of this type things often are not as simple as they first appear, owing to entropy differences between states of matter.
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