Physics: Kinetic Theory

Question

Kinetic theory

a. What is the average kinetic energy of a monatomic gas at a temperature of 354 K?

b. If this is Ne gas (neon has 20 nucleons) what is the root-mean-squared speed?

c. What temperature would be required to triple the speed of these particles?

Answer 1

a) If you mean per molecule, it's 3/2*kT, where k = Boltzmann's Constant, so that's 3/2(1.381*10^-23)(354) = 7.333*10^-21 J. For a mole of gas, multiply by Avogadro's Number, 6.022*10²³  = 4416 J/mol or 4.416 kJ/mol

 

b) V_rms = √(3RT/M)   R = Gas Constant; M = molar mass in kg. So V = √(3*8.314*354/0.02018) = 661.5 m/s

 

c) Since T is under a square root symbol, you would have to multiply it by 9 (because √9 = 3), so that's 354*9 = 3186 K

 

Answer 2

(a) For a single molecule of a mono-atomic gas, the average KE is

 

KE = (3/2) kT = (3/2)(1.38 x 10^-23 J/K)(354 K) = 7.328 x 10^-21 J

 

(b)

 

(1/2) * m * v_rms^2 = (3/2) kT

 

v_rms = sqrt(3kT / m)

 

m = m_Ne = 20 * (1.67 x 10^-27 kg) = 3.34 x 10^-26 kg  (i.e. about 20 proton masses)

 

v_rms = sqrt[3 * (1.38 x 10^-23 J/K)(354 K) / (3.34 x 10^-26 kg)] = 662.4 m/s

 

(c)  

 

3 v_rms = sqrt(3kT' / m)

 

Solve for T' and get

 

T' = 3 m v_rms^2 / k = 3 (3.34 x 10^-26 kg) (662.4 m/s)^2 / (1.38 x 10^-23 J/K) = 3186 K