Pressure of an ideal gas is obtained from kinetic gas equation. The kinetic gas equation is: PV=13(mNA)u2=13Mu2, where, mNA=M(molar mass)NA=Avogadro's numberu=root mean square velocityTranslational kinetic energy of n mole12Mu2=32(PV)=32(nRT)Average translational kinetic energy per molecule=32(RTNA)=32(KT)urms=√3PVM=√3RTM=√3Pdand uav=√8RTπMWhat is the rms speed of gas molecule of mass 10−12 gm at room temperature 27∘ C according to kinetic theory of gases?(Given R=8 JK−1mol−1, NA=6×1023)

Pressure of an ideal gas is obtained from kinetic gas equation. The ki

1.	Pressure of an ideal gas is obtained from kinetic gas equation. The kinetic gas equation is: PV=13(mNA)u2=13Mu2, where, mNA=M(molar mass)NA=Avogadro's numberu=root mean square velocityTranslational kinetic energy of n mole12Mu2=32(PV)=32(nRT)Average translational kinetic energy per molecule=32(RTNA)=32(KT)urms=√3PVM=√3RTM=√3Pdand uav=√8RTπMWhat is the rms speed of gas molecule of mass 10−12 gm at room temperature 27∘ C according to kinetic theory of gases?(Given R=8 JK−1mol−1, NA=6×1023)
Answer» Pressure of an ideal gas is obtained from kinetic gas equation. The kinetic gas equation is: $P V = \frac{1}{3} (m N_{A}) u^{2} = \frac{1}{3} M u^{2}$ , where, $m N_{A} = M$ (molar mass) $N_{A} = Avogadro's number$ $u = root mean square velocity$ $Translational kinetic energy of n mole \frac{1}{2} M u^{2} = \frac{3}{2} (P V) = \frac{3}{2} (n R T)$ $Average translational kinetic energy per molecule = \frac{3}{2} (\frac{R T}{N_{A}}) = \frac{3}{2} (K T)$ $u_{r m s} = \sqrt{\frac{3 P V}{M}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 P}{d}}$ and $u_{a v} = \sqrt{\frac{8 R T}{π M}}$ What is the rms speed of gas molecule of mass $10^{- 12}$ gm at room temperature $27^{\circ} C$ according to kinetic theory of gases? $(Given R = 8 J K^{- 1} m o l^{- 1}, N_{A} = 6 \times 10^{23})$

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