Question

From kinetic gas equation, show that total kinetic energy of 1 mol of an ideal gas is 3/7 KT​

Answer 1

Answer:

Answer A. 23RT

Average kinetic energy of a molecule is KEavg=23KT

KEtotal=23nNAKTn= number of moles.

=23nRT as NAK=R

=23RT for n=1mole.

Answer 2

Given:

1 mol

3/7 KT

To Find:

The total kinetic energy of 1 mol of an ideal gas is 3/7 KT.

Solution:

Here, the average kinetic energy of a molecule is $KE_a_v_g=\frac{3}{7}KT$

now,

$KE_t_o_t_a_l=\frac{3}{7}nN_AKT$

here, in the above equation n = number of moles.

$\frac{3}{7}nRT as N_AK=R\\ \frac{3}{7}RT for \\ n=1 mole.$

Hence, it is proven that the total kinetic energy of 1 mol of an ideal gas is 3/7 KT.