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

Answer:

The ratio of the pressure at the top to the pressure at the bottom is $\dfrac{701}{1000}$

Explanation:

Given that,

Number of molecules $n= 10^24$

Mass $m= 3\times10^{-26}\ kg$

Temperature = 300 K

Height $h = 5\times10^{3}$

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

Using barometric formula

$P_{h}=P_{0}e^{\dfrac{-mgh}{kT}}$

$\dfrac{P_{h}}{P_{0}}=e^{\dfrac{-mgh}{kT}}$

Where, m = mass

g = acceleration due to gravity

h = height

k = Boltzmann constant

T = temperature

Put the value in to the formula

$\dfrac{P_{h}}{P_{0}}=e^{\dfrac{-3\times10^{-26}\times9.8\times5\times10^{3}}{1.3807\times10^{-23}\times300}}$

$\dfrac{P_{h}}{P_{0}}=\dfrac{701}{1000}$

Hence, The ratio of the pressure at the top to the pressure at the bottom is $\dfrac{701}{1000}$

Answer 2

Answer:

Top pressure : Bottom pressure = 701 : 1000

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