Question

A sealed container of volume V = 0.10 m3 holds a sample of N = 3.0 × 1024 atoms of helium gas in equilibrium. The distribution of speeds of the helium atoms shows a peak at 1.1 × 103 m s−1. (i) Calculate the temperature and pressure of the helium gas. (ii) What is the average kinetic energy of the helium atoms? (iii) What is the energy corresponding to the maximum in the energy distribution? (Take the mass of each helium atom to be 4.0 amu.)

Answer 1

Given :

Volume (V) = 0.10 m³

N = 3.0 * 10²⁴ atoms (1mole = 6.022×10²³ atoms) = 4.98 moles

velocity = 1.1 * 10³ m/s

mass = 4.0 amu (1amu =1.6605e-27kg)

solution :

E=1/2 mv² (kinetic energy equation)

=1/2 *6.64*10^-27 * (1.1*10³)² = 4.02*10^-21 J

Now E = 3/2 KT

           where K is Boltzmann Constant =1.38*10^-23 J/K

So for T = 2E/3K

             =2*4.02*10^-21 J / 3*1.38*10^-23 J/K

             = 194.20 K

PV = nRT

so P = nRT /V

         = 4.98 * 8.314*194.20 / 0.1

         = 80434.01 pascal = 0.79 atm