### 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 The ratio of specific heats gamma$\gamma$ is related to degrees of freedom f$f$ by: gamma = fracC_PC_V = 1 + frac2f$\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$ ### Core Logic Determine the degrees of freedom $ ### Core Logic Determine the degrees of freedom $f for each type of gas: - **Monoatomic gas**: Translational only $ for each type of gas: - **Monoatomic gas**: Translational only $\implies f = 3 $ $gamma = 1 + frac23 = frac53 quad text(Matches C-I)$\gamma = 1 + \frac{2}{3} = \frac{5}{3} \quad \text{(Matches C-I)}$ - **Diatomic rigid gas**: Translational (3) + Rotational (2) $ - **Diatomic rigid gas**: Translational (3) + Rotational (2) $\implies f = 5 $ $gamma = 1 + frac25 = frac75 quad text(Matches D-II)$\gamma = 1 + \frac{2}{5} = \frac{7}{5} \quad \text{(Matches D-II)}$ ### Step 1: Check Remaining Categories - **Diatomic non-rigid gas**: Translational (3) + Rotational (2) + Vibrational (2) $ ### Step 1: Check Remaining Categories - **Diatomic non-rigid gas**: Translational (3) + Rotational (2) + Vibrational (2) $\implies f = 7 $ $gamma = 1 + frac27 = frac97 quad text(Matches B-IV)$\gamma = 1 + \frac{2}{7} = \frac{9}{7} \quad \text{(Matches B-IV)}$ - **Triatomic rigid gas**: Translational (3) + Rotational (3) $ - **Triatomic rigid gas**: Translational (3) + Rotational (3) $\implies f = 6 $ $gamma = 1 + frac26 = 1 + frac13 = frac43 quad text(Matches A-III)$\gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \quad \text{(Matches A-III)}$ This yields the matching order: A-III, B-IV, C-I, D-II. ### Pattern Recognition Sees: Degrees of freedom and $ This yields the matching order: A-III, B-IV, C-I, D-II. ### Pattern Recognition Sees: Degrees of freedom and $\gamma values. Shortcut: Lower degrees of freedom result in higher $ values. Shortcut: Lower degrees of freedom result in higher $\gamma values. Order of degrees of freedom: Monoatomic (3) < Diatomic rigid (5) < Triatomic rigid (6) < Diatomic non-rigid (7). Corresponding $ values. Order of degrees of freedom: Monoatomic (3) < Diatomic rigid (5) < Triatomic rigid (6) < Diatomic non-rigid (7). Corresponding $\gamma: $: $\frac{5}{3} > \frac{7}{5} > \frac{4}{3} > \frac{9}{7}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory

### Related Formula mathrmV_mathrmrms = sqrtfrac3mathrmRmathrmTmathrmM implies mathrmV_mathrmrms^2 = frac3mathrmRmathrmMmathrmT$\mathrm{V}_{\mathrm{rms}} = \sqrt{\frac{3\mathrm{R}\mathrm{T}}{\mathrm{M}}} \implies \mathrm{V}_{\mathrm{rms}}^2 = \frac{3\mathrm{R}}{\mathrm{M}}\mathrm{T}$ ### Core Logic The parameter asked is the mean squared velocity, which corresponds directly to mathrmV_mathrmrms^2$\mathrm{V}_{\mathrm{rms}}^2$. From the ideal gas kinematics relation, we observe: mathrmV_mathrmrms^2 propto mathrmT$\mathrm{V}_{\mathrm{rms}}^2 \propto \mathrm{T}$ Comparing this format against standard geometric linear templates (y = mx$y = mx$), the curve must map as a clean straight line originating from absolute zero zero coordinates. ### Step 1: Final Conclusion This linear profile matches Graph (1), selecting option (1). ### Pattern Recognition Watch the vertical ordinate labels carefully: Root-mean-square velocity scales as a sub-linear curve (sqrtmathrmT$\sqrt{\mathrm{T}}$), while mean squared metric trends linearly directly (y propto x$y \propto x$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory