What Is the Logic Behind Adding Pressure in the Derivation of PV=nRT?

  • Thread starter Thread starter kidsasd987
  • Start date Start date
  • Tags Tags
    Derivation Pv=nrt
AI Thread Summary
The discussion centers on the derivation of the ideal gas law, PV=nRT, particularly the role of pressure in this equation. It clarifies that pressure (P) is defined as force (F) per unit area (A), indicating uniform pressure across a surface. The reasoning behind summing pressures from small cubes is explained as calculating the total pressure exerted on the interior of the cube by individual particles, rather than simply aggregating pressures from the cubes themselves. This approach emphasizes that the pressure in one cube reflects the overall pressure in the container when particle velocities are consistent. Understanding this logic is crucial for grasping the derivation of the ideal gas law.
kidsasd987
Messages
142
Reaction score
4
http://quantumfreak.com/derivation-of-pvnrt-the-equation-of-ideal-gas/please check eq.(7)

pressure equation is P=F/A which means, in any region over the surface, pressure will be the same.
for example, if we assume all the particles have the same mean squared velocity, pressure of one cube will be the same as the pressure of the whole container.

I wonder why we add up all the pressure we found for the small cubes, and what's the logic behind it?
 
Physics news on Phys.org
You're not adding up the pressure of small cubes, you're adding up the pressure exerted on the cube's interior per particle to get the total pressure.
 
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (First part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8 and stuck at some statements. It's little bit confused. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. Solution : The surface bound charge on the ##xy## plane is of opposite sign to ##q##, so the force will be...
Back
Top