A container of volume 10^{-3} m^3 contains 3 | vedclass

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Question

(A) Given: Volume $V = 10^{-3} \ m^3$,Number of molecules $N = 3.0 	imes 10^{22}$,Mass of one molecule $m = 5.3 	imes 10^{-26} \ kg$,$rms$ speed $v_

Vedclass · Accepted Answer

$8.48 	imes 10^4 \ N/m^2$

Vedclass · Answer

(A) Given: Volume $V = 10^{-3} \ m^3$,Number of molecules $N = 3.0 	imes 10^{22}$,Mass of one molecule $m = 5.3 	imes 10^{-26} \ kg$,$rms$ speed $v_{rms} = 400 \ m/s$.The formula for pressure exerted by an ideal gas is given by $P = \frac{1}{3} \frac{N m}{V} v_{rms}^2$.Substituting the values:$P = \frac{1}{3} 	imes \frac{(3.0 	imes 10^{22}) 	imes (5.3 	imes 10^{-26})}{10^{-3}} 	imes (400)^2$$P = \frac{1}{3} 	imes \frac{15.9 	imes 10^{-4}}{10^{-3}} 	imes 160000$$P = 5.3 	imes 10^{-1} 	imes 160000$$P = 0.53 	imes 160000 = 84800 \ N/m^2 = 8.48 	imes 10^4 \ N/m^2$.

$A$ container of volume $10^{-3} \ m^3$ contains $3.0 \times 10^{22}$ gas molecules,each having a mass of $5.3 \times 10^{-26} \ kg$. If the $rms$ speed of the molecules is $400 \ m/s$,calculate the pressure exerted by the gas.

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