Q. Two kg of a monoatomic gas is at a pressure of $4 \times 10^4 \; N/m^2$ . The density of the gas is $8 \; kg /m^3$. What is the order of energy of the gas due to its thermal motion ?

Solution:

Thermal energy of N molecule
$ = N\left(\frac{3}{2}kT\right) $
$ =\frac{N}{N_{A}} \frac{3}{2}RT $
$ = \frac{3}{2}\left(nRT\right) $
$ =\frac{3}{2}PV $
$=\frac{3}{2}P\left(\frac{m}{8}\right)$
$ = \frac{3}{2} \times4\times10^{4} \times\frac{2}{8} $
$ =1.5 \times10^{4} $
order will $10^4$

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions

5. The spectrum of an oil flame is an example for ...........

KCET 2010 Dual Nature Of Radiation And Matter