Question

RMS velocity of so2 at 400k is x cm/sec, now RMS velocity of ch4 at 900k is​

Answer 1

Given:

RMS velocity of so2 at 400K is x cm/sec.

To find:

RMS velocity of CH4 at 900K ?

Calculation:

For Sulphur Dioxide:

$\therefore \: x = \sqrt{ \dfrac{3RT}{m} }$

$\implies\: x = \sqrt{ \dfrac{3RT}{32 + 32} }$

$\implies\: x = \sqrt{ \dfrac{3R(400)}{64} }$

$\implies\: x = \dfrac{(20) \sqrt{3R}}{8}$

$\implies\: \sqrt{3R} = \dfrac{8x}{20} \: . \: . \: . \: .(1)$

Now, for Methane (CH4):

$\therefore \: v_{rms} = \sqrt{\dfrac{3RT_{2} }{m_{2}} }$

$\implies \: v_{rms} = \sqrt{\dfrac{3R(900)}{12 + 4} }$

$\implies \: v_{rms} = \sqrt{\dfrac{3R(900)}{16} }$

$\implies \: v_{rms} = \dfrac{ (30)\sqrt{3R} }{4}$

$\implies \: v_{rms} = \dfrac{ 30 \times 8x}{4 \times 20}$

$\implies \: v_{rms} = 3x$

So, RMS velocity of methane is 3x

Answer 2

QUESTION

RMS velocity of SO2 at 400 k is x cm/sec , now RMS velocity of CH4 at 900 k is

SOLUTION

RMS velocity - The RMS speed ( root mean square speed ) I'd defined as the square root , of the mean of the square , of the random speeds of the individual molecule of a gas .

${ \boxed{ \tt{ \green{ \: V\: rms \: = \sqrt{ \frac{3RT}{M} } }}}}$

given that RMS velocity of SO 2 = x cm/sec

at temperature = 400 k

molecular mass of SO2 = 64g/mol

temperature ofr CH 4 = 900 K

molecular mass of CH4 = 16 g/mol

We have to find the RMS velocity of CH4 at 900 k

${ \tt{ \green{V \: rms \: SO2 = \sqrt{ \frac{3 \times R \times 400}{64}} = x \: cm \: {sec}^{ - 1} }} }$

${ \tt{ \green{V \: rms \: CH4 \: = \sqrt{ \frac{3 \times R \times 900}{16}}}} }$

dividing equation 2 by 1

${ \tt{ \blue{ \frac{V \: rms \: CH4}{x} = \frac{ \sqrt{ \frac{3 \times R \times 900}{16} } }{ \sqrt{ \frac{3 \times R \times 400}{64}}}} } }$

${ \rightarrow{ \tt{ \blue{ \: \frac{V \: rms \: CH4}{x} = \sqrt{ \frac{64 \times 900}{16 \times 400}}}}} }$

${ \rightarrow{ \tt{ \blue{ \: V\: rms \: CH4 \: = x \sqrt{9} \: = \: 3x}}}}$

RMS velocity of CH4 at 900 K = 3x cm / sec