physics 十萬火急! 20分!

Tiana

2010-09-12 1:19 am

A narrow glass tube, sealed at one end, has a column of airenclosed by a thread of mercury120 mm long. When the tube is held with its open end uppermost, the air column is 300 mm long. When the tube is inverted, the air column is 420mm long. Given that the density of mercury 1.36 x 10^4 kg m^-3. Calculate

a) the atmospheric pressure,
b) the length of air column when the tube is held horizontally.

更新1:

我係中5學生未教mmHg 呢樣老師說: Pressure of gas inside column x area = mg of mercury + atmospheric pressure ( pressure outside) x area 但我不明白

回答 (1)

天同

2010-09-12 1:41 am

✔ 最佳答案

(a) Let the atmospheric pressure be Po
Pressure of air column when the tube is upright = (Po + 120) mmHg
Pressure of sir column when the tube is inverted = (Po - 120) mmHg
Using Boyle's Law,
(Po + 120) x 300A = (Po-120) x 420A
where A is the cross-sectional area of the air column
hence, Po = 720 mmHg = (720/1000) x 1.36x10^4 x 9,81 N/m^2
= 9.6 x 10^4 N/m^2
[where 9.81 m/s^2 is the acceleration due to gravity g]

(b) When the tube is held horizontally, the pressure of the air in the tube is at atmospheric pressure Po.
Using Boyle' Law again,
Po x L.A = (Po +120) x 300A
where L is the length of air column
hence, L = 350 mm

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