Question

Molecules in a gas are at 18 degrees celsius. If the rms velocity of the molecules was increased by 4.7%, what temperature would the molecules be raised to?

Answer 1

$V$_rms$= \sqrt\frac{3RT}{M}\times \sqrt T$

$= \alpha T$^1/2 $(\alpha = \sqrt\frac{3R}{M})$

$\Delta V$_rms$= \frac{4.7 Vrms}{100}$

$\frac{4.7}{100} = 2 \times T\times \frac{4.7}{100}$

$\Delta T = \frac{9.4 T}{100}$ -----------------------------1

$\Delta T = \frac{9.4\times 291.5}{100}$

$\Delta T =27.361$ ----------------------------2

Final temperature will be

$T$_f $= T$₂$+ \Delta T$

$= 18+27.368 =45.368 C$