As the temperature is raised from 20 oC to 40 oC , the average kinetic energy of neon atoms changes by a factor of which of the following?

Question

As the temperature is raised from 20 oC to 40 oC , the average kinetic energy of neon atoms changes by a factor of which of the following? (A) (1/2) (B) 2 (C) (√313/293) (D) (313/293)

Tardigrade Admin · Accepted Answer

As average K.E. =(3/2)(R T/NA) i.e., textK. textE. propto textT Hence (K . E .313/K . E .293)=(313/293)

Q. As the temperature is raised from $20^{o} C$ to $40^{o} C$ , the average kinetic energy of neon atoms changes by a factor of which of the following?

$\frac{1}{2}$

$2$

$\frac{313}{293}$

$\frac{313}{293}$

As average K.E. $= \frac{3}{2} \frac{RT}{N _{A}}$
i.e., $K . E. \propto T$
Hence $\frac{K . E . _{313}}{K . E . _{293}} = \frac{313}{293}$

Q. As the temperature is raised from 20oC to 40oC , the average kinetic energy of neon atoms changes by a factor of which of the following?

21​

2

293313​​

293313​

As average K.E. =23​NA​RT​ i.e., K.E.∝T Hence K.E.293​K.E.313​​=293313​

Q. As the temperature is raised from $20^{o} C$ to $40^{o} C$ , the average kinetic energy of neon atoms changes by a factor of which of the following?

$\frac{1}{2}$

$2$

$\frac{313}{293}$

$\frac{313}{293}$

As average K.E. $= \frac{3}{2} \frac{RT}{N _{A}}$
i.e., $K . E. \propto T$
Hence $\frac{K . E . _{313}}{K . E . _{293}} = \frac{313}{293}$