Match the List-I with List-II

List-I	List-II
A. Triatomic rigid gas	I. fracC_PC_V=frac53$\frac{C_{P}}{C_{V}}=\frac{5}{3}$
B. Diatomic non-rigid gas	II. fracC_PC_V=frac75$\frac{C_{P}}{C_{V}}=\frac{7}{5}$
C. Monoatomic gas	III. fracC_PC_V=frac43$\frac{C_{P}}{C_{V}}=\frac{4}{3}$
D. Diatomic rigid gas	IV. fracC_PC_V=frac97$\frac{C_{P}}{C_{V}}=\frac{9}{7}$

Choose the correct answer from the options given below:

### Related Formula Ideal Gas Law: n = fracPVRT$$n = \frac{PV}{RT}$$ Conservation of moles: n_1 + n_2 = n_f$n_1 + n_2 = n_f$ ### Core Logic Let the volume of the smaller vessel be V_1 = V$V_1 = V$, then the volume of the larger vessel is V_2 = 2V$V_2 = 2V$. Initial moles in large vessel: n_2 = frac8 times 2VR times 1000 = frac16V1000R$$n_2 = \frac{8 \times 2V}{R \times 1000} = \frac{16V}{1000R}$$ Initial moles in small vessel: n_1 = frac7 times VR times 500 = frac14V1000R$$n_1 = \frac{7 \times V}{R \times 500} = \frac{14V}{1000R}$$ Total total initial moles: n_texttotal = n_1 + n_2 = frac30V1000R$$n_{\text{total}} = n_1 + n_2 = \frac{30V}{1000R}$$ ### Step 1: Connect Vessels to Dynamic Equilibrium When connected, the total final volume is V_f = V + 2V = 3V$V_f = V + 2V = 3V$. The final temperature is T_f = 600text K$T_f = 600\text{ K}$. Using mole conservation: frac30V1000R = fracP_f (3V)R times 600$$\frac{30V}{1000R} = \frac{P_f (3V)}{R \times 600}$$ frac301000 = frac3P_f600 implies frac301000 = fracP_f200$$\frac{30}{1000} = \frac{3P_f}{600} \implies \frac{30}{1000} = \frac{P_f}{200}$$ P_f = frac30 times 2001000 = 6text kPa$$P_f = \frac{30 \times 200}{1000} = 6\text{ kPa}$$

### Pattern Recognition Connecting chambers preserves the net mass/moles (sum n_i = textconstant$\sum n_i = \text{constant}$). Keep everything relative to a common volume multiplier V$V$ to easily cancel terms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory

### Related Formula f = frac1T = fracv_textavglambda$$f = \frac{1}{T} = \frac{v_{\text{avg}}}{\lambda}$$ where: * f$f$ = frequency of collisions * v_textavg$v_{\text{avg}}$ = average speed of the molecules * lambda$\lambda$ = mean free path ### Core Logic Given parameters: * Average speed, v_textavg = 600mathrm~m/s$v_{\text{avg}} = 600\mathrm{~m/s}$ * Mean free path, lambda = 3 times 10^-7mathrm~m$\lambda = 3 \times 10^{-7}\mathrm{~m}$ ### Step 1: Calculate Frequency Substitute the values into the formula: f = frac6003 times 10^-7 = 2 times 10^9mathrm~s^-1$$f = \frac{600}{3 \times 10^{-7}} = 2 \times 10^{9}\mathrm{~s}^{-1}$$ Hence, the collision frequency is 2 times 10^9/mathrms$2 \times 10^{9}/\mathrm{s}$. ### Pattern Recognition Collision frequency is simply distance covered per unit time (average velocity) divided by the average distance between consecutive collisions (mean free path). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory

### 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

### Related Formula The total translational kinetic energy of a gas sample depends only on the number of moles and absolute temperature, regardless of whether the molecule is monoatomic or polyatomic: K.E._texttranslational = frac32 n R T$$K.E._{\text{translational}} = \frac{3}{2} n R T$$ ### Core Logic Given data: * Mass of mathrmCO_2$\mathrm{CO}_2$ gas, m = 50 text g$m = 50 \text{ g}$ * Molar mass of mathrmCO_2$\mathrm{CO}_2$, M = 44 text g/mol$M = 44 \text{ g/mol}$ * Absolute Temperature, T = 17 + 273.15 = 290.15 text K$T = 17 + 273.15 = 290.15 \text{ K}$ * Universal gas constant, R approx 8.314 text J/(molcdottextK)$R \approx 8.314 \text{ J/(mol}\cdot\text{K)}$ Calculate total moles n$n$: n = frac5044 approx 1.1364 text moles$$n = \frac{50}{44} \approx 1.1364 \text{ moles}$$ Substitute values into the expression : K.E._texttranslational = frac32 times frac5044 times 8.314 times 290.15$$K.E._{\text{translational}} = \frac{3}{2} \times \frac{50}{44} \times 8.314 \times 290.15$$ K.E._texttranslational = 1.5 times 1.1364 times 8.314 times 290.15 approx 4108.6 text J$$K.E._{\text{translational}} = 1.5 \times 1.1364 \times 8.314 \times 290.15 \approx 4108.6 \text{ J}$$ The closest matching value specified in the test alternatives is 4102.8 mathrmJ$4102.8 \mathrm{J}$. ### Pattern Recognition A common mistake is using frac52nRT$\frac{5}{2}nRT$ or frac72nRT$\frac{7}{2}nRT$ because mathrmCO_2$\mathrm{CO}_2$ is a triatomic linear molecule. Remember that **translational** kinetic energy is always frac32nRT$\frac{3}{2}nRT$ for any gas sample, as translation has exactly 3 degrees of freedom regardless of the molecular layout. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases