Physics Problems & Solutions: Thermal Physics

Using kinetic theory analysis, show that pressure of ideal gas is proportional to the average translational kinetic energy and number density.

$P=\frac{2}{3} \times \frac {1}{2} m\langle v^2\rangle \times \frac {N}{V}$

Solution:

Force: $F=ma=m \frac{\Delta v}{\Delta t}=\frac{\Delta p} {\Delta t}$

Total Force: $F=n\frac{\Delta p} {\Delta t}$

n = Number of collision in time Δt
Number of particles that moves in one direction highly likely is half of total particle, n = ½ N

Density: N/V
Volume of container : V = v_x ∆tA

$n= \frac{1}{2} N \frac{V}{V}=\frac{1}{2} \frac {N}{V} v_x \Delta tA$

Δp = Momentum transferred to wall per elastic collision.
Ideal gas → elastic collision → particle moves in x-direction with velocity v_x and bounces back with velocity −v_x.

$\Delta p=m[v_x-(-v_x ) ]=2mv_x$

Force:

$F= \left(\frac{1}{2} \frac {N}{V} v_x \Delta tA \right )\left(\frac{2mv_x }{\Delta t}\right )$

Pressure:

$P=\frac{F}{A}= \frac{\left(\frac{1}{2} \frac {N}{V} v_x \Delta tA \right )\left(2mv_x\right )}{\Delta tA}=\frac {N}{V}mv^2_x$

Average velocity, equal probability:

$\langle v^2 \rangle=\langle v_x^2 \rangle+\langle v_y^2 \rangle+\langle v_z^2 \rangle$
$\langle v_x^2 \rangle=\langle v_y^2 \rangle=\langle v_z^2 \rangle=\frac{1}{3}\langle v^2 \rangle$

Pressure:

$P=\frac{1}{3}\frac {N}{V}m\langle v^2\rangle$

In terms of average kinetic energy and number density, pressure:

$P=\frac{2}{3} \times \frac {1}{2} m\langle v^2\rangle \times \frac {N}{V}$

Pressure of ideal gas is proportional to the average translational kinetic energy and number density

Physics Problems & Solutions

Thermal Physics - Kinetic Theory Analysis

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