Question

The root mean square of gas molecules at 25 K and 1.5 bar is 100 $ms^{- 1}$ . 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)

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

So $\frac{100 \, m / s}{u_{r m s}^{′}}=\sqrt{\frac{25 \text{ K}}{100 \text{ K}}}=\frac{1}{2}$

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

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