A Challenge!

sl_ye

2007-05-18 8:23 pm

1. A weather balloon floats up into the atmosphere. The air pressure decreases as the balloon floats higher. Assuming the temperature remains constant, the balloon:
a) increases in volume
b) decreases in volume
c) maintains a steady height in the moving air currents
d) maintains a constant volume
e) remains in the stratosphere

更新1:

2. A sample of gas is collected at 25C. If the temperature of the gas is tripled and the pressure on the gas is doubled, what portion of the original volume of gas will remain?

更新2:

3. A mixture of gases in a cylinder contains 0.85mol of CH4, 0.55mol of oxygen, 1.25mol of nitrogen, and 0.27mol of propane (C3H8). The pressure on the cylinder gauge reads 2573kPa. What pressure does each gas exert in the cylinder?

回答 (1)

魏王將張遼

2007-05-18 10:21 pm

✔ 最佳答案

(1) Answer = C
Since as the balloon floats up into the atmosphere, the external pressure decreases which makes the balloon increases in volume since initially, the air inside the balloon is of atmospheric pressure on the ground.
Eventually, as the balloon's height increases, the air density outside it will decreases which means that, the thrust on the balloon by the air will be smaller (by calculation, the decrease in thrust due to decrease in air density is more significant than the increase in thrust due to the increase in balloon's volume) and so, finally, the thrust and weight of the balloon will balance out each other and then it remains in a steady height.
(2) Suppose the "triple" means in the value in Celsius, then final temperature will be 75°C. Then by the general gas law:
P' V' / T' = PV/T
With P' = 2P, T' = 348 K and T = 298 K:
V' = 348PV/(248 × 2P)
= 0.702 V
(3) By Dalton's law: Partial pressure is proportional to mole fraction, we have:
Mole fractino of CH4 = 0.291
Mole fraction of N2 = 0.428
Mole fraction of O2 = 0.188
Mole fraction = propane = 0.092
So, partial pressures are:
CH4 = 749 kPa
N2 = 1101 kPa
O2 = 485 kPa
Propane = 298 kPa

參考: My physics knowledge

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