### Related Formula 1. Ideal Gas Law: P V = n R T$P V = n R T$ 2. Internal Energy (U$U$) of a diatomic gas (f = 5$f = 5$ degrees of freedom): U = n C_v T = n left(frac52 Rright) T = frac52 P V$U = n C_v T = n \left(\frac{5}{2} R\right) T = \frac{5}{2} P V$ ### Core Logic Given parameters: - Dimensions of the room = 4 \ mathrmm times 4 \ mathrmm times 3 \ mathrmm$4 \ \mathrm{m} \times 4 \ \mathrm{m} \times 3 \ \mathrm{m}$ - Volume of air in the room V = 4 times 4 times 3 = 48 \ mathrmm^3$V = 4 \times 4 \times 3 = 48 \ \mathrm{m}^3$ - Room pressure P = 1 \ mathrmatm = 10^5 \ mathrmN/m^2$P = 1 \ \mathrm{atm} = 10^5 \ \mathrm{N/m^2}$ ### Step 1: Calculate internal energy Using the thermodynamic relationship: U = frac52 P V$U = \frac{5}{2} P V$ Substitute the volume and pressure parameters: U = frac52 times 10^5 \ mathrmN/m^2 times 48 \ mathrmm^3$U = \frac{5}{2} \times 10^5 \ \mathrm{N/m^2} \times 48 \ \mathrm{m}^3$ U = 5 times 10^5 times 24 = 120 times 10^5 \ mathrmJ = 12 times 10^6 \ mathrmJ$U = 5 \times 10^5 \times 24 = 120 \times 10^5 \ \mathrm{J} = 12 \times 10^6 \ \mathrm{J}$ Thus, the internal energy is 12 times 10^6 mathrm~J$12 \times 10^6 \mathrm{~J}$. ### Pattern Recognition Sees: Internal energy of diatomic gas occupying a macroscopic room volume. Trap: Attempting to calculate thermodynamic variables like temperature or density explicitly. Internal energy is completely determined by pressure and volume via degrees of freedom (U = fracf2PV$U = \frac{f}{2}PV$). Shortcut: A diatomic gas has f=5$f=5$, so U = 2.5 PV$U = 2.5 PV$. Putting in numbers, U = 2.5 times 10^5 times 48 = 120 times 10^5 = 12 times 10^6 mathrm~J$U = 2.5 \times 10^5 \times 48 = 120 \times 10^5 = 12 \times 10^6 \mathrm{~J}$, leaving a coefficient of 12. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula gamma = 1 + frac2f$\gamma = 1 + \frac{2}{f}$ ### Core Logic Let's find the specific heat ratio for each gas based on degrees of freedom: 1. **Monoatomic gas A:** - Degrees of freedom, f_A = 3$f_A = 3$ (translational only) gamma_A = 1 + frac23 = frac53$\gamma_A = 1 + \frac{2}{3} = \frac{5}{3}$ 2. **Polyatomic gas B:** - Translational degrees of freedom = 3 - Rotational degrees of freedom = 3 - Vibrational modes = 1. *Note: Each active vibrational mode has 2 degrees of freedom (kinetic + potential energy terms).* This contributes 2 times 1 = 2$2 \times 1 = 2$ degrees of freedom. - Therefore, the total active degrees of freedom is: f_B = 3 + 3 + 2 = 8$f_B = 3 + 3 + 2 = 8$ The specific heat ratio of B is: gamma_B = 1 + frac2f_B = 1 + frac28 = 1 + frac14 = frac54$\gamma_B = 1 + \frac{2}{f_B} = 1 + \frac{2}{8} = 1 + \frac{1}{4} = \frac{5}{4}$ Now, find the ratio of specific heat capacities: fracgamma_Agamma_B = frac5/35/4 = frac43$\frac{\gamma_A}{\gamma_B} = \frac{5/3}{5/4} = \frac{4}{3}$ We are given: fracgamma_Agamma_B = 1 + frac1n implies frac43 = 1 + frac1n implies frac1n = frac13 implies n = 3$\frac{\gamma_A}{\gamma_B} = 1 + \frac{1}{n} \implies \frac{4}{3} = 1 + \frac{1}{n} \implies \frac{1}{n} = \frac{1}{3} \implies n = 3$ ### Step 1: Final Conclusion The value of n$n$ is 3. ### Pattern Recognition Always remember that each vibrational mode contributes exactly 2 degrees of freedom because it holds both kinetic and potential energy components (f_textvib = 2 times textmodes$f_{\text{vib}} = 2 \times \text{modes}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

### Related Formula For an ideal gas in a closed container, the volume V$V$ remains constant (isochoric process). By Gay-Lussac's Law: P propto T Rightarrow fracDelta PP = fracDelta TT$P \propto T \Rightarrow \frac{\Delta P}{P} = \frac{\Delta T}{T}$ where T$T$ must be in Kelvin. ### Core Logic Given parameters: - Percent increase in pressure: fracDelta PP times 100 = 0.4\% Rightarrow fracDelta PP = 0.004$\frac{\Delta P}{P} \times 100 = 0.4\% \Rightarrow \frac{\Delta P}{P} = 0.004$ - Increase in temperature Delta T = 1^circmathrmC = 1mathrm~K$\Delta T = 1^{\circ}\mathrm{C} = 1\mathrm{~K}$ ### Step 1: Calculate initial temperature (T$T$) Substitute the values into Gay-Lussac's fractional variance formula: 0.004 = frac1T$0.004 = \frac{1}{T}$ T = frac10.004 = 250mathrm~K$T = \frac{1}{0.004} = 250\mathrm{~K}$ ### Pattern Recognition A standard percentage-increase layout. A change of 0.4\%$0.4\%$ means frac1250$\frac{1}{250}$ of the original quantity. Hence, a 1mathrm~K$1\mathrm{~K}$ rise corresponds to an initial temperature of 250mathrm~K$250\mathrm{~K}$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula textK.E. = fracf2 k_B T$\text{K.E.} = \frac{f}{2} k_B T$ [cite: 688] ### Core Logic The average kinetic energy per molecule depends only on the temperature T$T$ and the degrees of freedom f$f$ of the gas[cite: 75, 688]. Both Helium (mathrmHe$\mathrm{He}$) and Argon (mathrmAr$\mathrm{Ar}$) are monoatomic noble gases, meaning both share the same degrees of freedom (f = 3$f = 3$)[cite: 689]. Since they sit in the same flask at identical room temperature (T = 300\ textK$T = 300\ \text{K}$), their translational kinetic energies per molecule are exactly equal [cite: 74, 688]: fractextK.E._mathrmHetextK.E._mathrmAr = frac11$\frac{\text{K.E.}_{\mathrm{He}}}{\text{K.E.}_{\mathrm{Ar}}} = \frac{1}{1}$ [cite: 688] ### Pattern Recognition Do not get distracted by the molar masses given in the question stem[cite: 76, 77]. Kinetic energy per molecule is purely temperature-dependent for an ideal gas, unlike the root-mean-square velocity (v_textrms$v_{\text{rms}}$) which explicitly includes molecular weight. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula The root-mean-square (r.m.s.) velocity of gas molecules is given by: v_textrms = sqrtfrac3RTM$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ Since molecular weight M$M$ is directly proportional to the vapour density (rho$\rho$), the r.m.s. velocity is inversely proportional to the square root of its vapour density: fracv_textrms1v_textrms2 = sqrtfracrho_2rho_1$\frac{v_{\text{rms1}}}{v_{\text{rms2}}} = \sqrt{\frac{\rho_2}{\rho_1}}$ ### Core Logic Given the ratio of vapour densities : fracrho_1rho_2 = frac425$\frac{\rho_1}{\rho_2} = \frac{4}{25}$ Therefore, the ratio of their r.m.s. velocities is: fracv_textrms1v_textrms2 = sqrtfrac254 = frac52$\frac{v_{\text{rms1}}}{v_{\text{rms2}}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$ ### Pattern Recognition R.M.S. velocity changes inversely with the square root of mass or density. Whenever a density ratio is given, simply invert the fraction and take the square root to immediately find the velocity ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases