Question

1) Modern surface science research is often carried out in ultra-high vacuum (UHV) chambers that can typically reach pressures on the order of 10-11 torr.
a. How many gas molecules are there in each cubic centimeter of a UHV chamber at this pressure at room temperature?
b. What is the chamber’s molar volume?
c. The average density of interstellar space is roughly 1 H atom/cm3 and its temperature is typically on the order of 10 K. What are the pressure and molar volume of this interstellar medium?

Answer 1

a)

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Temperature

R = ideal gas constant

PV= nRT

V = 1 cm3 = 1 mL = 10^-3 L

T = 25°C = 298 K

P = 10 torr,

n = PV/(RT) = (10)(10^-3)/(62.4*298)

n = 5.37773*10^-7 moles

1 mol = 6.022*10^23 paticles

no particles = (5.37773*10^-7)(6.022*10^23) = 3.23846*10^17 particles per cm3

b)

chamber molar volume

V = volume / mol

V = (1cm3)/( 5.37773*10^-7 moles) = 1859520.65 cm3 per mol

c)

Density = 1 Hatom/cm3

T = 10K

a) Pressure.

PV = nRT

n = mass/MW

PV = mass/MW*RT

mass/V = D

P = D/MW*RT

Density = mass/volume = 1 atom of H --> 1/(6.022*10^23) g / cm3 = 1.6605*10^-24 g/cm3

change to g/L --> (1.6605*10^-24) g/cm3 * 1000 cm3 / L = 1.6605*10^-21 g/L

P = D/MW*RT

P = (1.6605*10^-21)/(1) * (62.4*10)

P = 1.0361*10^-18 torr

Molar V:

6.022*10^23 atoms = 1 mol of H

1 atom of H = x

x = 1/(6.022*10^23) moles

x = 1.660577*10^-24 moles

now...

Density = 1 atom / cm3

Density = ( 1.660577*10^-24 ) / cm3

D =  1.660577*10^-24 mol / cm3

we need mola V

inver D

V = 1/D = 1/( 1.660577*10^-24) = 6.0220*10^23 cm3 / mol