Physics Problems & Solutions: Pressure

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Thermal Physics - Isothermal

Suppose one mole of an ideal gas undergoes the reversible cycle ABCA shown in the P-V diagram above, where AB is an isotherm. The molar heat capacities are C_p at constant pressure and C_v at constant volume. The net heat added to the gas during the cycle is equal to

A. RT_h (V₂/V₁)
B. −C_p(T_h − T_c)
C. C_p(T_h − T_c)
D. RT_h ln (V₂/V₁) − C_p(T_h − T_c)
E. RT_h ln (V₂/V₁) − R(T_h − T_c)

(GR9677 #15)

Solution:

AB Isotherm → T Constant = T_h

Ideal Gas: PV = nRT
P = nRT / V

n = 1 mole,

W_AB = _{_V₁}∫^{^V₂}P dV = RT_h _{_V₁}∫^{^V₂}(1/V) dV = RT_h ln (V₂/V₁)

BC Isobaric → P constant = P₂

W_BC = _{_V₂}∫^{^V₁}P dV = P₂ (V₁− V₂) = P₂V₁ − P₂V₂

From the diagram:

P₂V₁ = nRT_c

P₂V₂ = nRT_h

W_BC = nR(T_c − T_h) = R(T_c − T_h)

W_CA = 0 since V constant

Total W = W_AB+ W_BC = RT_h ln (V₂/V₁) + R(T_c − T_h)

or

W = RT_h ln (V₂/V₁) − R(T_h − T_c)

Answer: E

Thermal Physics - Maxwell-Boltzmann Statistics

Consider a system of N non-interacting particles confined in a volume V at a temperature such that the particles obey classical Boltzmann statistics. If the temperature is lowered to the point at which quantum effects become important, the pressure of the gas may differ depending on whether the particles are fermions or bosons. Let P_F be the pressure exerted by the particles if they are fermions. P_B be the pressure if they are bosons and, and P_C be the pressure the particles would exert if quantum effects are ignored. Which of the following is true?

(GR8677 #88)

Solution:

Ideal gas: $PV=NkT \rightarrow P=\frac{N}{V}kT$

Classical (Maxwell-Boltzmann Statistics): $E=\frac{3}{2}kT$

$P=\frac{N}{V}kT\rightarrow P_C = \frac{2}{3} \frac{N}{V} \frac{3}{2} kT=\frac{2}{3} \frac{NE}{V}$

For Bosons (photon):

$P=\frac{N}{V}kT\rightarrow P_B = \frac{1}{3} \frac{N}{V} 3 kT=\frac{1}{3} \frac{NE}{V}$

→

For Fermion: $E=\frac{3}{5}E_F$
E_F

= Fermi Energy, $E_F \gg kT$

For the same temperature:

Answer: B

Thermal Physics - Pressure of photon gas

Compute the pressure exerted by gas of photons.

Solution:

According to kinetic theory analysis, pressure:

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

Momentum: p = mv
For photon, v = c
Energy of photon: E = pc

→ Pressure:

$P=\frac{1}{3}\frac {N}{V}pv=\frac{1}{3}\frac {N}{V}pc=\frac{1}{3}\frac {NE}{V}$

Ideal Gas: PV = NkT

→ Pressure:

$PV=\frac{1}{3}NE\rightarrow NkT =\frac{1}{3}NE$

→ Energy:

Thermal Physics - Kinetic Theory Analysis

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