Kinetic Theory and PV Diagrams

Here you can see we used the definition of velocity to find the velocity in one direction for  a single molecule. Then we can use the definition of F=dp/dt=mv-(-mv)/dt=2mv/dt, We the can double the time dtx= x/Vx(the time to get to the wall) so it get back to its original location. Finally we can say Fx=mVx^2/X.

Now we can go ahead and say F=P/A, therfore P=F/A which means P=Fx/X^2=mVx^2/X^3.
Now if we want to assumer there are N molecules all we have to do is multiple by N. We can also add other directions of freedom if we let Vx^2=Vy^2=Vz^2 we can say  Vtotal^2=3Vx^2.  and finally we can rewrite our expression for Pressure as P=NmVtotal^2/3V, or even once more as P=2N(Ekin)/3V. (notice how we let mVtotal^2=2(Ekin))

Now from the equation up above  we can say PV=2N(Ekin)/3 which we know is also equal to NkbT. We set them equal and we can find the velocity of the molecules. We can also find the a new equation by understanding the Eint can be defined as N(number of paticles) multiplies by the kinetic energy of each particle. doing this we find that Eint=3/2NkbT. On the left you see and isothermal compression where we find Q=W and on the right you see and adiabatic compression (dv=0)
 Eint = -W => 3/2nKdT=-pdV we take the integral and we find the C=T^3/2V

Solving the change in internal energy, First we solve for the Work needed to raise Temperature to 50 degrees Celsius from 20 degrees Celsius of a copper bar. Then we solve for Q which is equal to mcdeltaT which we find to equal 11.58kJ and then knowing  the first law of thermodynamics we get deltaE=Q-W = 11.58*10^3 J.

Here we use the equation derived above except instead of T^3/2V we use T^5/2V, this is because oxygen is diatomic and not mono-atomic. We got Tfinal = 604K or 628 degrees Fahrenheit, which we can compare to the 451 degrees needed to burn paper, which we learned in English! So we can expect the paper to burn! Don't believe me? video evidence below!