If He(g) has an average kinetic energy of 6670 J/mol under certain conditions, what is the root mean square speed of Cl2(g) molecules under the same conditions?

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If He(g) has an average kinetic energy of 6670 J/mol under certain conditions, what is the root mean square speed of Cl2(g) molecules under the same conditions?

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Answer 1 · 2016-11-19T11:59:22

we know

$V_{rms} = \sqrt{\frac{R T}{M}}$ $V_"rms"=sqrt((RT)/M)$

where

$V_{rms} \to RMS velocity of the gas$ $V_"rms"->"RMS velocity of the gas"$

$T \to absolute temperature of the gas$ $T->"Absolute temperature of the gas"$

$M \to Molar mass of the gas$ $M->"Molar mass of the gas"$

$R \to Universal gas constant$ $R->"Universal gas constant"$

So average molar kinetic energy of the gas

$E = \frac{1}{2} M V_{rms}^{2} = \frac{1}{2} R T$ $E=1/2MV_"rms"^2=1/2RT$

This equation reveals that molar KE is independent of the nature of the gas . It only depends on temperature as ideal behavior is concerned. So both He(g) and $C l_{2} (g)$ $Cl_2(g)$ will have same average $K E = 6670 J/mol$ $KE=6670"J/mol"$ .under the same condition of temperature.

So for $C l_{2} (g)$ $Cl_2(g)$

$\frac{1}{2} M_{C l_{2} (g)} V_{r m s C l_{2}}^{2} = 6670$ $1/2M_(Cl_2(g))V_(rmsCl_2)^2=6670$

$Taking atomic mass of Cl = 35.5 g/mol = 35.5 \times 10^{- 3} kg/mol$ $color(red)("Taking atomic mass of Cl"=35.5"g/mol"=35.5xx10^-3"kg/mol")$

$\Rightarrow V_{r m s C l_{2} (g)}^{2} = \frac{6670 \times 2}{2 \times 35.5 \times 10^{- 3}}$ $=>V_(rmsCl_2(g))^2=(6670xx2)/(2xx35.5xx10^-3)$

$\Rightarrow V_{r m s C l_{2} (g)} = \sqrt{\frac{6670000}{35.5}} \approx 433.5 m s^{- 2}$ $=>V_(rmsCl_2(g))=sqrt(6670000/35.5)~~433.5ms^-2$

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If He(g) has an average kinetic energy of 6670 J/mol under certain conditions, what is the root mean square speed of Cl2(g) molecules under the same conditions?

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