# 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 ?

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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 $10^4$

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