Physics Questions-Gases 快40分
2011-04-30 12:34 pm
A balloon rises from the ground. It will burst when its volume 1.5times the original value. Assume that whenever the balloon rises for 1 km, thetemperature of air falls by 10°C andthe atmospheric pressure falls by 2 x 10^4 Pa.A) The temperature and pressure of air inside the balloon are 30°C and 1 x 10^5 Parespectively on the ground. Calculate the height above the ground that theballoon bursts. (1.87 m)B) Hence, find the pressure and temperature at the height found in (a) (11.3°C)
回答 (2)
2011-04-30 6:05 pm
✔ 最佳答案
A) Suppose that original vol. of the balloon is V, then when it bursts it is at a height of h km where:Volume = 1.5V (as per given)
Temp. = 303 - 10h
Pressure = (10 - 2h) x 104
Using general gas law:
105V/303 = 1.5V x (10 - 2h) x 104/(303 - 10h)
10/303 = (15 - 3h)/(303 - 10h)
3030 - 100h = 4545 - 909h
809h = 1515
h = 1.87
b) Pressure = (10 - 2h) x 104 = 6.26 x 104 Pa
Temp. = 30 - 1.87 x 10 = 11.3 C
參考: 原創答案
2011-04-30 6:09 pm
(a) Let the height be h.
Temperature at height h = [30 - 10h]'C
Pressure at height h = [10^5 - h(2x10^4)] Pa
Use ideal gas equation: PV/T = constant
(10^5).V/(273+30) = [10^5 - (h/1000)(2x10^4)].(1.5V)/(273+ [30 -(h/1000).10])
where V is the volume of the balloon at ground level
solve for h gives h = 1.87 km
(b) Pressure = [10^5 - 1.87 x (2x10^4)] Pa = 62 600 Pa
Temperature = (30 - 10x1.87)'C = 11.3'C
Temperature at height h = [30 - 10h]'C
Pressure at height h = [10^5 - h(2x10^4)] Pa
Use ideal gas equation: PV/T = constant
(10^5).V/(273+30) = [10^5 - (h/1000)(2x10^4)].(1.5V)/(273+ [30 -(h/1000).10])
where V is the volume of the balloon at ground level
solve for h gives h = 1.87 km
(b) Pressure = [10^5 - 1.87 x (2x10^4)] Pa = 62 600 Pa
Temperature = (30 - 10x1.87)'C = 11.3'C
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