The Boltzmann equation IB DP Physics Study Notes - 2025 Syllabus
The Boltzmann equation IB DP Physics Study Notes
The Boltzmann equation IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
that the change in momentum of particles due to collisions with a given surface gives rise to pressure in gases and, from that analysis, pressure is related to the average translational speed of molecules as given by $P=\frac{1}{3} \rho v^2$
the relationship between the internal energy $U$ of an ideal monatomic gas and the number of molecules or amount of substance as given by $U=\frac{3}{2} N k_{\mathrm{B}} T$ or $U=\frac{3}{2} R n T$
the temperature, pressure and density conditions under which an ideal gas is a good approximation of a real gas.
Standard level and higher level: 6 hours
Additional higher level: There is no additional higher level content
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Average kinetic/internal energy of an ideal gas
- Since ideal gases have no intermolecular forces, their internal energy is stored completely as kinetic energy.
- The individual molecules making up an ideal gas all travel at different speeds:
- Without proof, the average kinetic energy EK of each ideal gas molecule has the following forms:
- kB is called the Boltzmann constant
EXAMPLE:
2.50 moles of hydrogen gas is contained in a fixed volume of $1.25 \mathrm{~m}^3$ at a temperature of $175^{\circ} \mathrm{C}$.
a) What is the average kinetic energy of each atom?
b) What is the total internal energy of the gas?
▶️Answer/Explanation
SOLUTION:
$T(\mathrm{~K})=175+273=448 \mathrm{~K}$.
a) $\overline{E_K}=\left(\frac{3}{2}\right) k_B T=\left(\frac{3}{2}\right)\left(1.38 \times 10^{-23}\right)(448)=9.27 \times 10^{-21} \mathrm{~J}$.
b) From $n=\frac{N}{N_A}$ we get $N=n N_A$.
$N=(2.50 \mathrm{~mol})\left(\frac{6.02 \times 10^{23} \mathrm{atoms}}{m o l}\right)=1.51 \times 10^{24} \mathrm{~atm}$.
$E_K=N \overline{E_K}=\left(1.51 \times 10^{24}\right)\left(9.27 \times 10^{-21}\right)=14000 \mathrm{~J}$.
Internal energy
- Since the ideal gases have no intermolecular forces, their internal energy is stored completely as kinetic energy
- As seen in previous section the average kinetic energy of a gas particle is :
- If we want to know total internal energy of an ideal gas, we must multiply this number by the total number of particles
