Question

A gas X2Y4, at 35°C has RMS
speed 12ms^-1. On heating the
gas twice to the original absolute
temperature, the dimer totally
dissociated to give monomer.
What is the RMS speed of XY2
molecules at the given elevated
temperature?
​

Answer 1

Answer:

Initial RMS speed, U=

M

3RT

When temperature is doubled,

U

f

=

M/2

3R×(2T)

=2

M

3RT

=2U

Explanation:

Vrms =

M

3RT

i.e,V

rms

∝

T

Answer 2

The RMS Speed of $X(Y)_{2}$ is 24 m/s

Given,

RMS Speed = 12 m/s

Temperature = 35°C

To Find,

Find the RMS Speed of $X(Y)_{2}$ molecules at the given elevated temperature

Solution,

• The square root of the mean square speed of all molecules is known as the root mean square speed (vrms). It is represented by the notation vrms = v2
• RMS speed is directly related to square root of temperature and inversely proportional to square root of molecular mass. At a specific temperature, molecules with lighter masses often move more quickly than those with greater weights.
• At the same temperature, lighter molecules like hydrogen and helium have higher "vrms" than heavier molecules like oxygen and nitrogen.
• As the temperature rises, molecules move at a faster rate than before.

According to the question we have to find the RMS speed of $X(Y)_{2}$

Here,

RMS speed of XY is $C_{1}$ = 12 m/s

RMS speed of $X(Y)_{2}$ is $C_{2}$ = ?

$T_{1}$ = 35 °C = 35+ 273 = 308 K

$T_{2}$ = 308 × 2 = 616 K

because, we are heating the

gas twice to the original absolute

temperature

We know that,

$M_{1} = 2M_{2}$

$\frac{C_{2} }{C_{1} } = \sqrt{\frac{T_{2} }{T_{1} } * \frac{M_{1} }{M_{2} } }$

$= \sqrt{\frac{616}{308} * \frac{2M_{2} }{M_{2} } }$

$\sqrt{\frac{4}{1} }$

= 2:1

so, we found that,

=> $C_{2}$ = 2× $C_{1}$

=> $C_{2}$ = 2× 12 = 24

so, $C_{2}$ = 24 m/s

Hence, the RMS Speed of $X(Y)_{2}$ is 24 m/s

#SPJ3

For more information about RMS speed, you can refer to the following questions

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