At 3 text{ atm} pressure,the weight of 14.8 t | vedclass

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(C) Using the ideal gas equation $PV = \frac{w}{M}RT$,where $P = 3 	ext{ atm}$,$V = 14.8 	ext{ L}$,$w = 15 	ext{ g}$,and $R = 0.0821 	ext{ L atm K

Vedclass · Accepted Answer

$8.7 	imes 10^4 	ext{ cm/s}$

Vedclass · Answer

(C) Using the ideal gas equation $PV = \frac{w}{M}RT$,where $P = 3 	ext{ atm}$,$V = 14.8 	ext{ L}$,$w = 15 	ext{ g}$,and $R = 0.0821 	ext{ L atm K}^{-1} 	ext{ mol}^{-1}$.$3 	imes 14.8 = \frac{15}{M} 	imes 0.0821 	imes T$$\frac{T}{M} = \frac{3 	imes 14.8}{15 	imes 0.0821} \approx 36.05 	ext{ K g}^{-1} 	ext{ mol}$.The formula for average speed is $\overline{v} = \sqrt{\frac{8RT}{\pi M}}$.Substituting $R = 8.314 	imes 10^7 	ext{ erg K}^{-1} 	ext{ mol}^{-1}$ (for $CGS$ units) and $\frac{T}{M} = 36.05$:$\overline{v} = \sqrt{\frac{8 	imes 8.314 	imes 10^7 	imes 36.05}{3.14159}}$$\overline{v} = \sqrt{7.63 	imes 10^9} \approx 8.73 	imes 10^4 	ext{ cm/s}$.

At $3 \text{ atm}$ pressure,the weight of $14.8 \text{ L}$ of a gas is $15 \text{ g}$. What is the average speed of the gas molecules?

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