Question

For a monoatomic gas, kinetic energy = E. The relation with rms velocity is:

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Since $p=\frac{1}{3}\times \frac{m}{v}\times u_{rms}^2$, we have

$u_{rms}={\left(\frac{3pV}{m}\right)}^{\frac{1}{2}}$------------------1

where $u_{rms}$ is the rms velocity for a molecule . Substituting $pV=kT$ in $e{q}^n$ 1, we get

$u_{rms}={\left(\frac{3kT}{m}\right)}^{\frac{1}{2}}$------------------2

Therefore,the kinetic energy = $\frac{3}{2}$ kT or $2E$ = $3kT$. Substituting in $e{q}^n$ 2, we get

$u_{rms}={\left(\frac{2E}{m}\right)}^{\frac{1}{2}}$