13.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

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Payal Gupta | Contributor-Level 10

9 months ago

13.9 Temperature of the helium atom, $T_{H e}$ = – 20 °C = 253 K and temperature of argon atom be = $T_{A r}$

Atomic mass of helium, $M_{H e}$ = 4.0 u

Atomic mass of Argon, $M_{A r}$ = 39.9 u

Let ${V_{r m s}}_{A r}$ be the rms speed of Argon and ${V_{r m s}}_{H e}$ be the rms speed of Helium

From the relation of $V_{r m s} = \sqrt{\frac{3 k T}{M}}$ we get

rms speed of Argon, ${V_{r m s}}_{A r} = \sqrt{\frac{3 k T_{A r}}{M_{A r}}}$

rms speed of Helium, ${V_{r m s}}_{H e} = \sqrt{\frac{3 k T_{H e}}{M_{H e}}}$

Since both the s

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physics ncert solutions class 11th 2023

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