Question

The average translational kinetic energy of 10 gram hydrogen molecules (H2)at 27°c

Answer 1

$\large\underline{\bigstar \: \: {\sf Given-}}$

• 10g of hydrogen gas
• Temperature (T) = 27°C

$\large\underline{\bigstar \: \: {\sf To \: Find -}}$

• Average translational kinetic energy (K)

$\large\underline{\bigstar \: \: {\sf Formula \: Used -}}$

$\large\color{violet}\bullet\underline{\boxed{\sf K=\dfrac{3}{2}nRT}}$

Here,

K = Translation Kinetic Energy

n = moles of ${\sf H_2}$

R = Gas Constant = 8.314 ${\sf Jmol^{-1}K^{-1}}$

T = Temperature = 27 + 273 = 300 K

$\large\underline{\bigstar \: \: {\sf Solution-}}$

we know that

$\implies\underline{\boxed{\sf Moles\:(n)=\dfrac{Mass}{Molecular \: Mass}}}$

Molecular mass of ${\sf H_2=2}$

Mass = 10g (Given)

$\implies{\sf n=\dfrac{10}{2}}$

$\color{orange}\implies{\sf Mole\:(n)=5}$

$\implies{\sf K=\dfrac{3}{2}×5×8.314×300}$

$\implies{\sf K=\dfrac{37413}{2} }$

$\implies{\sf K=18706.5J }$

$\color{red}\implies{\sf K=18.7\:kJ}$

$\large\underline{\bigstar \: \: {\sf Answer-}}$

Average translational kinetic energy is $\color{red}{\sf 18.7\:kJ}$