Since according to equation (\ref{kin}) each particle carries (on average) a kinetic energy of \(\frac{3}{2}k_B T\), the total energy of all the particles and thus the energy of the gas can be obtained by multiplying the mean kinetic energy of a particle with the total number of particles \(N\). In order to connect the macroscopically observed state variables of a gas such as temperature, volume and pressure with the microscopic variables such as particle mass and particle velocity, the kinetic theory of gases was developed. The relationship between pressure and temperature can be explained using the kinetic theory of gases. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Solution: Example 2. the internal energy in the gas per unit volume! He touched the tyre and found that it was warm. In order to determine the gas pressure in the cylinder, the number of particles that collide with the piston within a certain time \(\Delta t\) must first be determined.

Problem: What is the relationship between the pressure and the temperature of a fixed mass of gas at constant volume?

Equation 13.3.2 will give us, at a temperature of 20 °C = 293.15 K, CP − CV (molar) = 728 J K−1 kmole−1 = 0.09R.

First, it is assumed that the gases are ideal gases.

On the other hand, the gas pressure depends on the speed at which the particles hit the piston surface.

Note that the speed \(v\) is no longer limited to the x-direction but represents the total speed of a particle (to be precise: to the mean of the speed squares of all molecules)!

This means in particular: In the article “Gas pressure“, the formation of the gas pressure has already been explained in detail using the particle model. Here are order-of-magnitude figures for copper at room temperature (for exact figures, we would have to specify the exact temperature).

The Pressure Law (Gay-Lussac’s Law) gives the relationship between the pressure and temperature of a fixed mass of gas at constant volume. The apparatus is set up as shown in Figure. β is negative), so that, if we add heat at constant pressure, work is done on the water by its surroundings, and hence (we might argue, though erroneously), for water at 2 ºC, CP < CV. What is the relationship between the mean energy of the molecules and the temperature? If equation (\ref{ppp}) is expanded with factor 2, then the term \(\frac{1}{2}m \overline{v_x^2}\) can be interpreted as the mean kinetic energy \(\overline{W_{kin,x}}\) of the particles that is related to the x-direction: \begin{align}& p =\frac{2}{2} \cdot \frac{N}{V} \cdot m \cdot \overline{v_x^2} = 2 \cdot \frac{N}{V} \cdot \overbrace{\frac{1}{2} m \cdot \overline{v_x^2}}^{\overline{W_{kin,x}}} \\[5px] \label{druck}& \boxed{p = 2 \cdot \frac{N}{V} \cdot \overline{W_{kin,x}}} \\[5px]\end{align}.

Problem: What is the relationship between the pressure and the temperature of a fixed mass of gas at constant volume?

Equation 13.3.2 will give us, at a temperature of 20 °C = 293.15 K, CP − CV (molar) = 728 J K−1 kmole−1 = 0.09R.

First, it is assumed that the gases are ideal gases.

On the other hand, the gas pressure depends on the speed at which the particles hit the piston surface.

Note that the speed \(v\) is no longer limited to the x-direction but represents the total speed of a particle (to be precise: to the mean of the speed squares of all molecules)!

This means in particular: In the article “Gas pressure“, the formation of the gas pressure has already been explained in detail using the particle model. Here are order-of-magnitude figures for copper at room temperature (for exact figures, we would have to specify the exact temperature).

The Pressure Law (Gay-Lussac’s Law) gives the relationship between the pressure and temperature of a fixed mass of gas at constant volume. The apparatus is set up as shown in Figure. β is negative), so that, if we add heat at constant pressure, work is done on the water by its surroundings, and hence (we might argue, though erroneously), for water at 2 ºC, CP < CV. What is the relationship between the mean energy of the molecules and the temperature? If equation (\ref{ppp}) is expanded with factor 2, then the term \(\frac{1}{2}m \overline{v_x^2}\) can be interpreted as the mean kinetic energy \(\overline{W_{kin,x}}\) of the particles that is related to the x-direction: \begin{align}& p =\frac{2}{2} \cdot \frac{N}{V} \cdot m \cdot \overline{v_x^2} = 2 \cdot \frac{N}{V} \cdot \overbrace{\frac{1}{2} m \cdot \overline{v_x^2}}^{\overline{W_{kin,x}}} \\[5px] \label{druck}& \boxed{p = 2 \cdot \frac{N}{V} \cdot \overline{W_{kin,x}}} \\[5px]\end{align}.