Calculate the kinetic energy (i) per mole (ii) per unit mass(iii) pe

1.	Calculate the kinetic energy (i) per mole (ii) per unit mass(iii) per molecule of nitrogen at STP. [Molar mass of nitrogen = 28 × 10-3 kg/mol, R = 8.314 J/mol.K, kB = 1.38 × 10-23 J/K]
Answer» Data : T = 273 K, M₀ = 28 × 10^-3 kg/mol, R = 8.314 J/mol.K, k_B = 1.38 × 10^-23 J/K (i) The KE per mole = \(\frac32RT\) = \(\frac32\)(8.314)(273) = 3.404 × 10³ J/mol (ii) The KE per unit mass = \(\frac32\)\(\frac{RT}{M_0}\) = \(\frac32\). \(\frac{(8.314)(273)}{28\times10^{-3}}\) = 1.216 × 10⁵ J/kg (iii) The KE per molecule = \(\frac32\)k_BT = \(\frac32\). (1.38 x 10^-23)(273) = 5.651 x 10^-21)

Calculate the kinetic energy (i) per mole (ii) per unit mass(iii) per molecule of nitrogen at STP. [Molar mass of nitrogen = 28 × 10-3 kg/mol, R = 8.314 J/mol.K, kB = 1.38 × 10-23 J/K]

Answer»

Data : T = 273 K, M₀ = 28 × 10^-3 kg/mol, R = 8.314 J/mol.K, k_B = 1.38 × 10^-23 J/K

(i) The KE per mole = \(\frac32RT\)

= \(\frac32\)(8.314)(273) = 3.404 × 10³ J/mol

(ii) The KE per unit mass = \(\frac32\)\(\frac{RT}{M_0}\)

= \(\frac32\). \(\frac{(8.314)(273)}{28\times10^{-3}}\) = 1.216 × 10⁵ J/kg

(iii) The KE per molecule

= \(\frac32\)k_BT

= \(\frac32\). (1.38 x 10^-23)(273) = 5.651 x 10^-21)