At T K temperature,the average speed of O_2 w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The average speed of $O_2$ is given by $V_{av} = \sqrt{\frac{8RT}{\pi M_{O_2}}}$,where $M_{O_2} = 32 	imes 10^{-3} \, kg/mol$.The $rms$ speed of

Vedclass · Accepted Answer

$1320$

Vedclass · Answer

(A) The average speed of $O_2$ is given by $V_{av} = \sqrt{\frac{8RT}{\pi M_{O_2}}}$,where $M_{O_2} = 32 	imes 10^{-3} \, kg/mol$.The $rms$ speed of $N_2$ is given by $V_{rms} = \sqrt{\frac{3RT'}{M_{N_2}}}$,where $T' = 47 + 273 = 320 \, K$ and $M_{N_2} = 28 	imes 10^{-3} \, kg/mol$.According to the problem: $\sqrt{\frac{8RT}{\pi 	imes 32}} = \frac{7}{4} \sqrt{\frac{3R 	imes 320}{28}}$.Squaring both sides: $\frac{8RT}{32\pi} = \frac{49}{16} 	imes \frac{3R 	imes 320}{28}$.Simplifying the equation: $\frac{RT}{4\pi} = \frac{49 	imes 3R 	imes 320}{16 	imes 28}$.Solving for $T$: $T = \frac{49 	imes 3 	imes 320 	imes 4\pi}{16 	imes 28} \approx 1320 \, K$ (taking $\pi \approx \frac{22}{7}$ for calculation).

At $T$ $K$ temperature,the average speed of $O_2$ will be $\left( \frac{7}{4} \right)$ of the $rms$ speed of $N_2$ at $47\,^oC$. Find the value of $T$.

Similar Questions

The rms speed of oxygen at room temperature is about $500 \,m/s$. The rms speed of hydrogen at the same temperature is about (in $\,m/s$)

If the pressure of a gas is doubled while keeping the temperature constant,what will happen to the root mean square $(RMS)$ speed of the gas molecules?

The $rms$ speed of a gas at room temperature of $27^{\circ}C$ is found to be $412 \ m/s$. The gas is:

The rms velocity of a gas molecule of mass $m$ at a given temperature is proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module