derive an expression for the pressure of an ideal gas in a container from kinetic energy
Answers
The time between collisions with the wall is the distance of travel between wall collisions divided by the speed.
1. t=2Lvx
The frequency of collisions with the wall in collisions per second is
2. f=1t=12L/vx=vx2L
According to Newton, force is the time rate of change of the momentum
3. F=dpdt=ma
The momentum change is equal to the momentum after collision minus the momentum before collision. Since we consider the momentum after collision to be mv, the momentum before collision should be in opposite direction and therefore equal to –mv.
4. Δp=mvx−(−mvx)=2mvx
According to equation #3, force is the change in momentum Δp divided by change in time Δt. To get an equation of average force F¯¯¯¯ in term of particle velocity vx, we take change in momemtum Δp multiply by the frequency f from equation #2.
5. F¯¯¯¯=Δp(f)=2mv(vx2L)=mv2xL
The pressure, P, exerted by a single molecule is the average force per unit area, A. Also V=AL which is the volume of the rectangular box.
6. P1Molecule=F¯¯¯¯A=(mv2xL)/A=mv2xLA=mv2xV
Let’s say that we have N molecules of gas traveling on the x-axis. The pressure will be
7. PNMolecules=mV(v2x1+v2x2+v2x3….+v2xN)=∑Na=0mv2xaV
To simplify the situation we will take the mean square speed of N number of molecules instead of summing up individual molecules. Therefore, equation #7 will become
8. PNParticles=Nmv2x¯¯¯¯¯V
Earlier we are trying to simplify the situation by only considering that a molecule with mass m is traveling on the x axis. However, the real world is much more complicated than that. To make a more accurate derivation we need to account all 3 possible components of the particle’s speed, vx, vy and vz.
9. v2¯¯¯¯¯=v2x¯¯¯¯¯+v2y¯¯¯¯¯+v2z¯¯¯¯¯
Since there are a large number of molecules we can assume that there are equal numbers of molecules moving in each of co-ordinate directions.
10. v2x¯¯¯¯¯=v2y¯¯¯¯¯=v2z¯¯¯¯¯
Because the molecules are free too move in three dimensions, they will hit the walls in one of the three dimensions one third as often. Our final pressure equation becomes
11. P=Nmv2¯¯¯¯¯3V
However to simplify the equation further, we define the temperature, T, as a measure of thermal motion of gas particles because temperature is much easier to measure than the speed of the particle. The only energy involve in this model is kinetic energy and this kinetic enery is proportional to the temperature T.
12. Ekinetic=mv22∝T
To combine the equation #11 and #12 we solve kinetic energy equation #12 for mv2.
13. mv2=2Ekinetic⇒mv23=2Ekinetic3
Since the temperature can be obtained easily with simple daily measurement like a thermometer, we will now replace the result of kinetic equation #13 with with a constant R times the temperature, T. Again, since T is proportional to the kinentic energy it is logical to say that T times k is equal to the kinetic energy E. k, however, will currently remains unknown.
14. kT=mv23=2Ekinetic3
Combining equation #14 with #11, we get:
15. P=NVmv2¯¯¯¯¯3=NV2Ekinetic3=NVkT=NkTV
Because a molecule is too small and therefore impractical we will take the number of molecules, N and divide it by the Avogadro’s number, NA= 6.0221 x 1023/mol to get n (the number of moles)
16. n=NNa
Since N is divided by Na, k must be multiply by Na to preserve the original equation. Therefore, the constant R is created.
17. R=Nak
Now we can achieve the final equation by replacing N (number of melecules) with n (number of moles) and k with R.
17. P=nRTV⇒PV=nRT
Calculation of R & k
According to numerous tests and observations, one mole of gas is a 22.4 liter vessel at 273K exerts a pressure of 1.00 atmosphere (atm). From the ideal gas equation above:
A. R=PVnT
B. R=(1atm)(22.4L)(1mol)(273K)=0.082LatmmolK
C. k=RNa⇒k=0.082Latm/molK6.0221x1023/mol=1.3806504x10−