Category: HC Verma Part 2: Heat & Thermo
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HCV Ch24 P10 – Kinetic Theory: Temperature at Which RMS Speed Equals Escape Velocity
Problem Statement Solve the gravitation problem: At what temperature is the rms speed of hydrogen molecules equal to the escape velocity from Earth (11.2 km/s)? ($M_{H_2} = 2 \times 10^{-3}$ kg/mol, $R = 8.314$ J/mol·K) $v_{rms} = v_e = 11200$ m/s $M = 2 \times 10^{-3}$ kg/mol $$v_{rms} = \sqrt{\frac{3RT}{M}} \Rightarrow T = \frac{M v_{rms}^2}{3R}…
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HCV Ch24 P9 – Kinetic Theory: Comparison of RMS Speeds of Different Gases
Problem Statement Solve the kinematics problem: At the same temperature, compare the rms speeds of hydrogen (H₂) and oxygen (O₂) molecules. $M_{H_2} = 2$ g/mol, $M_{O_2} = 32$ g/mol Same temperature $T$ $$v_{rms} = \sqrt{\frac{3RT}{M}} \Rightarrow v_{rms} \propto \frac{1}{\sqrt{M}}$$ So the ratio of speeds: $$\frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac Given Information Initial velocity $u$ (or $v_0$) Acceleration…
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HCV Ch24 P8 – Kinetic Theory: Number Density from Ideal Gas Law
Problem Statement Solve the thermodynamics problem: Calculate the number of molecules per cm³ in an ideal gas at 0°C and 1 atm (Loschmidt’s number). ($R = 8.314$ J/mol·K, $N_A = 6.022 \times 10^{23}$ mol$^{-1}$) $T = 273.15$ K, $P = 1.013 \times 10^5$ Pa $R = 8.314$ J/mol·K, $N_A = 6.022 \times 10^{23}$ mol$^{-1}$ From…
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HCV Ch24 P7 – Maxwell-Boltzmann Speed Distribution: Most Probable Speed
Problem Statement Solve the kinematics problem: Find the most probable speed of nitrogen molecules at 300 K. ($M = 28 \times 10^{-3}$ kg/mol, $R = 8.314$ J/mol·K) $T = 300$ K $M = 28 \times 10^{-3}$ kg/mol (N₂) The Maxwell-Boltzmann distribution has three characteristic speeds: $$v_p = \sqrt{\frac{2RT}{M}} \quad (\text{most probable})$$ $$v_{avg} Given Information Initial…
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HCV Ch24 P6 – Degrees of Freedom and Internal Energy of Ideal Gas
Problem Statement Solve the thermodynamics problem: Find the total internal energy of 1 mole of diatomic gas (like O₂ or N₂) at 300 K. $n = 1$ mol, $T = 300$ K Diatomic gas: $f = 5$ degrees of freedom (3 translational + 2 rotational) $R = 8.314$ J/mol·K By the equipartition theorem, each degree…
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HCV Ch24 P5 – Kinetic Theory: Pressure from Molecular Impacts
Problem Statement Solve the fluid mechanics problem: Calculate the pressure exerted by $10^{23}$ nitrogen molecules in a 1-litre container at 300 K. ($m_{N_2} = 4.65 \times 10^{-26}$ kg, $k_B = 1.38 \times 10^{-23}$ J/K) $N = 10^{23}$, $V = 1$ L $= 10^{-3}$ m³ $T = 300$ K $m = 4.65 \times 10^{-26}$ kg per…
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HCV Ch24 P4 – Kinetic Theory: Mean Free Path of Gas Molecules
Problem Statement Estimate the mean free path of nitrogen molecules at 300 K and 1 atm pressure. (Diameter of N₂ molecule $d = 3.7 \times 10^{-10}$ m, $P = 1.01 \times 10^5$ Pa) Given Information All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section) Physical Concepts…
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HCV Ch24 P3 – Ideal Gas Law: Finding Pressure
Problem Statement Solve the thermodynamics problem: A gas occupies 0.5 m³ at 300 K and contains $6.02 \times 10^{25}$ molecules. Find the pressure. ($k_B = 1.38 \times 10^{-23}$ J/K) $V = 0.5$ m³ $N = 6.02 \times 10^{25}$ molecules $T = 300$ K $k_B = 1.38 \times 10^{-23}$ J/K The ideal gas law in terms…
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HCV Ch24 P2 – Kinetic Theory: Average Kinetic Energy per Molecule
Problem Statement Solve the work-energy problem: Find the average kinetic energy of a gas molecule at 300 K. ($k_B = 1.38 \times 10^{-23}$ J/K) $T = 300$ K $k_B = 1.38 \times 10^{-23}$ J/K The equipartition theorem gives average kinetic energy per molecule (3 translational degrees of freedom): $$\langle KE \rangle = \frac{3}{2}k_B T$$ Step…
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HCV Ch24 P1 – Kinetic Theory: RMS Speed of Gas Molecules
Problem Statement Solve the kinematics problem: Find the rms speed of hydrogen molecules at 300 K. ($M_{H_2} = 2 \times 10^{-3}$ kg/mol, $R = 8.314$ J/mol·K) $T = 300$ K $M = 2 \times 10^{-3}$ kg/mol (molar mass of H₂) $R = 8.314$ J/mol·K The rms speed comes from equating kinetic energy to thermal energy:…