At what temperature and 1 atm pressure will | vedclass

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(A) The root mean square speed $(u_{rms})$ is given by the formula $u_{rms} = \sqrt{\frac{3RT}{M}}$.
For $CO_2$ at $STP$,$T = 273.15 \ K$ and $M = 44

Vedclass · Accepted Answer

$- 99.27 \ ^\circ C$

Vedclass · Answer

(A) The root mean square speed $(u_{rms})$ is given by the formula $u_{rms} = \sqrt{\frac{3RT}{M}}$.
For $CO_2$ at $STP$,$T = 273.15 \ K$ and $M = 44 \ g/mol$.
For $N_2$,$M = 28 \ g/mol$. We need to find $T$ such that $u_{rms}(N_2) = u_{rms}(CO_2)$.
Equating the expressions: $\sqrt{\frac{3RT_{N_2}}{M_{N_2}}} = \sqrt{\frac{3RT_{CO_2}}{M_{CO_2}}}$.
This simplifies to $\frac{T_{N_2}}{M_{N_2}} = \frac{T_{CO_2}}{M_{CO_2}}$.
Substituting the values: $\frac{T_{N_2}}{28} = \frac{273.15}{44}$.
$T_{N_2} = \frac{273.15 	imes 28}{44} \approx 173.83 \ K$.
Converting to Celsius: $T(^\circ C) = 173.83 - 273.15 = -99.32 \ ^\circ C$. The closest option is $-99.27 \ ^\circ C$.

At what temperature and $1 \ atm$ pressure will the root mean square speed of $N_2$ gas be equal to the root mean square speed of $CO_2$ gas at $STP$?

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