obtain an expression for the pressure of ideal gas
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Answer:
XPRESSION FOR THE PRESSURE OF A GAS: Let us suppose that a gas is enclosed in a cubical box having length l. Let there are 'N' identical molecules each having mass 'm'. Since the molecules are of same mass and perfectly elastic, so their mutual collisions result in the interchange of velocities only. Only collisions with the walls of the container contribute to the pressure by the gas molecules. Let us focus on a molecule haivig velocity v1 and components of velocity vx1,yy1,vz1 along x,y and z-axis as shown in figure
(##RESHATPHYXIC01E01006_A01##)
v21=v2x1+v2y1+v2z1
The change in momentum of the molecule after one collision with wall BCGF
=mvx1−(−mvx1)=2mvx1.
The time taken between the successive impacts
on the face BCGF=distancevelocity=2lvx1
Time rate of change of momentum due to
collision =chan≥∈momentumtimetaken
=2mvx12l/vx1=mv2x1l
Hence the net force on the wall BCGF due to the impact of n molecules of the gas is:
Fx=mv2x1l+mv2x2l+mv2x3l+........+mv2xnl=ml
(v2x1+v2x2+v2x3+......+v2xn)=mNl<v2x>
where <v2x>= mean square velocity in x-direction. Since molecules do not favour any particular direction therefore <v2x>=<v2y>
=<v2z>.But <v2>
=<v2x>+<v2y>+<v2z>
⇒<v2x>=<v2>3. Pressure is equal to force divided by area.
P=Fxl2=M3l3<v2>=M3V<v2>. Pressure is independent of x,y,z diretions.
Where l3= volume of the container =V
M= total mass of the gas, <v2>= mean square speed of molecules
⇒P=13ρ<v2>
of gas =12M<v2>=32PV=32nRT
Translational kinetic energy of 1 molecule
=32kT (it is independent of nature of gas)
<v2>=3Pρorvrms=3Pρ−−−√=3RTMMole−−−−−−√=3kTm−−−−√
Where vrms is root mean square speed of teh gas.
Pressure exerted by the gas is P=13ρ<v2>
=23×12ρ<v2>orP=23E,E=32P
Explanation: