Question

The root mean square of gas molecules at 25 K and 1.5 bar is 100 $m{s}^{-1}$ . If the temperature is raised to 100 K and the pressure to 6.0 bar, the root mean square speed becomes

• A. $200\text{ m}{\text{s}}^{-1}$
• B. $100\text{ m}{\text{s}}^{-1}$
• C. $400\text{ m}{\text{s}}^{-1}$
• D. $1600\text{ m}{\text{s}}^{-1}$

Answer 1

Answer: (A)

$\frac{{\text{u}}_{\text{rms}}}{{\text{u}}_{\text{rms}}^{{}^{\prime }}}=\sqrt{\frac{\frac{3\text{RT}}{\text{M}}}{\frac{3\text{RT’}}{\text{M}}}}=\sqrt{\frac{\text{T}}{\text{T’}}}$

So $\frac{100m/s}{u_{rms}^{\prime }}=\sqrt{\frac{25\text{ K}}{100\text{ K}}}=\frac{1}{2}$

or $u_{rms}^{\prime }=2\times 100\text{ m}{\text{s}}^{-1}$

$=200\text{ m}{\text{s}}^{-1}$ .