Question

A gas consists of 1024 molecules, each with mass 3 × 10-26 kg. It is heated to a temperature of 300 K, while the volume is held constant. 1) If the gas is confined to a vertical tube 5 × 103 m high, what is the ratio of the pressure at the top to the pressure at the bottom?

Answer 1

Top pressure : Bottom pressure = 701 : 1000

Step-by-step explanation:

Number of molecules = n = 10^24

Height = h = 5 × 10^3 m

Mass = m = 3 × 10^-26 kg

Boltzman’s Constant = K = 1.38 × 10^-23 J/K

Temperature = T = 300K

The formula for barometer pressure is given Below:

Ph = P0 e^-(mgh/KT)

Ph/P0 = e^-(3 × 10^-26 × 9.81 × 5 × 10^3)/(1.38 × 10^-23)(300)

Ph/P0 = e^-0.355

Ph/P0 = 1/e^0.355

Ph/p0 =0.7008 = 700.8/1000 = 701/1000

Hence,

Top pressure : Bottom pressure = 701 : 1000

Answer 2

The ratio of the pressure at the top to the pressure at the bottom is
`(701)/(1000)`

Step-by-step explanation:

Given that,

Number of molecules
`n= 10^24`

Mass
`m= 3*10^(-26)\ kg`

Temperature = 300 K

Height
`h = 5*10^(3)`

We need to calculate the ratio of the pressure at the top to the pressure at the bottom

Using barometric formula

`P_(h)=P_(0)e^{(-mgh)/(kT)}`

`(P_(h))/(P_(0))=e^{(-mgh)/(kT)}`

Where, m = mass

g = acceleration due to gravity

h = height

k = Boltzmann constant

T = temperature

Put the value in to the formula

`(P_(h))/(P_(0))=e^{(-3*10^(-26)*9.8*5*10^(3))/(1.3807*10^(-23)*300)}`

`(P_(h))/(P_(0))=(701)/(1000)`

Hence, The ratio of the pressure at the top to the pressure at the bottom is
`(701)/(1000)`