Question

The total kinetic energy of 0.6 mol
of an ideal gas at 27 C is?

Answer 1

3/2 NRT formulla is used and answer would be 1122.3

Answer 2

Answer:

Total Kinetic energy of 0.6 moles of gas is 2.245 kilo Joules.

Explanation:

Average kinetic energy is defined as the average of the kinetic energies of all the particles present in a system. It is determined by the equation:

$K.E.=\frac{3RT}{2N_A}$

where,

K.E = Average kinetic energy  of a molecule of a gas

R = Gas constant

T = Temperature of the system

$N_A$ = Avogadro's number =$6.022\times 10^{23} mol^{-1}$

T = 27°C = 300.15 K

$K.E.=\frac{3\times 8.314 J/mol K\times 300.15 K}{2\times 6.022\times 10^{23} mol^{-1}}=6.214\times 10^{-21} J$

In 0.6 moles = $0.6 mol\times 6.022\times 10^{23} mol^{-1}$ molecules

Total Kinetic energy of 0.6 moles of gas :

$K.E\times 0.6 mol\times 6.022\times 10^{23} mol^{-1}$

$=6.214\times 10^{-21} J\times 0.6 mol\times 6.022\times 10^{23} mol^{-1}$

Total Kinetic energy  = 2,245.15 Joules = 2.245 kiloJoules