Calculate the total and average kinetic energy of 32 g of methane mole

1.	Calculate the total and average kinetic energy of 32 g of methane molecules at 27ºc.
Answer» Average K.E. = \(\frac{3}{2}\)K_bT = \(\frac{3}{2}\)\(\frac{RT}{N_A}\) Rs = 8∙314 JK^-1 mol^-1 , N_A = 6∙022 × 10²³ T = 27 + 273 = 300 K K.E = \(\frac{3}{2}\) x \(\frac{8.314\times300}{6.022\times10^{23}}\) = 6.22 x 10^-21 J Total К.Е. = \(\frac{3}{2}\)nRT = \(\frac{3}{2}\) x \(\frac{32}{16}\) x 8.314 × 300 = 7482.6 J

Calculate the total and average kinetic energy of 32 g of methane molecules at 27ºc.

Answer»

Average K.E. = \(\frac{3}{2}\)K_bT

= \(\frac{3}{2}\)\(\frac{RT}{N_A}\)

Rs = 8∙314 JK^-1 mol^-1 , N_A = 6∙022 × 10²³

T = 27 + 273 = 300 K

K.E = \(\frac{3}{2}\) x \(\frac{8.314\times300}{6.022\times10^{23}}\)

= 6.22 x 10^-21 J

Total К.Е. = \(\frac{3}{2}\)nRT

= \(\frac{3}{2}\) x \(\frac{32}{16}\) x 8.314 × 300 = 7482.6 J