物理-氣體分子運動論問題(2)

I put down my alternative solution just to show that there's more than 1 way to tackle with this problem.
先假設如 天同所言, 份子質量為2.0E-26 kg .


假設容器的體積為y m^3
則氣體的總質量為0.5y kg
份子數量 = 0.5y / 2.0e-26 = 2.5y e25 粒
==> 2.5y e25 / 6.022e23 = 41.51y 摩爾(mole)

然後代入理想氣體方程:
PV = nRT
(P) x (y) = (41.51y) x 8.314 x (273+30)
y = 104570 Pa ~ 1x10^5 Pa

毋須用到玻爾茲曼常數.
如果你堅持份子質量的單位是g, 那答案就乘1000 好了.

以上算法是單純用物理化學關於氣體的部份強行拆解; 不過, 反正理想氣體方程都是由氣體粒子運動論衍生出來, 答案是肯定一樣的了. 差在是否用到微觀角度.

Since kinetic energy of a gas molecule = (1/2)mc^2 = (3/2)kT
where m is the mass of a molecule
c^2 is the mean-square speed of the molecule
k is Boltzmann constant
T is the absolute temperature

Hence, c^2 = 3kT/m = 3 x (1.38x10^-23) x (273+30)/(2x10^-26) (m/s)^2
= 6.27x10^5 (m/s)^2

Pressure = (1/3) x 0.5 x 6.27x10^5 Pa = 10^5 Pa

[Note: the mass of molecule given in your question should be of order of 10^-26 kg, 10^-26 g is quite a small mass for a molecule].