HKDSE practice paper

code PLUS

2012-06-07 3:17 am

A fixed mass of an ideal gas is contained in a cylinder fitted with a frictionless piston. If the gas is cooled under constant pressure,

(1) the average separation of the gas molecules will decrease.
(2) the r.m.s. speed of the gas molecules will decrease.
(3) the number of collisions per second of the gas molecules on the piston will decrease.

A. (1) and (2) only
B. (1) and (3) only
C. (2) and (3) only
D. (1), (2) and (3)

A stone falls from rest. Neglecting air resistance, the ratio of the distance travelled by the stone in the 1st second to that travelled in the 2nd second is

A. 1 : 1
B. 1 : 2
C. 1 : 3
D. 1 : 4

回答 (1)

天同

2012-06-07 3:57 am

✔ 最佳答案

1. Statement (1) is correct.
When the gas is cooled, its volume decreases (Charles' Law). This indicates the molecular separation decreases.

Statement (2) is correct.
The r.m.s. speed of molecules of a given gas depends on temperature. The higher the temperature, the higher is the r.m.s. speed. A decrease of temperature thus leads to a decrease of r.m.s. speed.

Statement (3) is wrong.
The no. of collisions per second of gas molecules gives rise to pressure. Since the pressure is kept constant, this implies that the no. of collisions per second of gas molecules doesn't change.

2. Use equation of motion: s = ut + (1/2)at^2
For the first second, u = 0, a = g, t = 1 s, s = s1
hence, s1 = (1/2)g

For the first two seconds, u = 0, a = g, t = 2 s, s = s2
hence, s2 = (1/2)g(2^2) = 2g
Displacement for the 2nd second s' = s2 - s1 = 2g - g/2 = 3g/2

Therefore, s1/s' = (g/2)/(3g/2) = (g/2) x (2/3g) = 1:3

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