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 V_textrms = sqrtfrac3RTM$$V_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$ ### Core Logic Given that the RMS speed of oxygen molecules equals the RMS speed of hydrogen molecules: V_textrms(O_2) = V_textrms(H_2)$$V_{\text{rms}(O_2)} = V_{\text{rms}(H_2)}$$ sqrtfrac3RT_O_2M_O_2 = sqrtfrac3RT_H_2M_H_2$$\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}$$ Squaring both sides: fracT_O_2M_O_2 = fracT_H_2M_H_2$$\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$$ ### Step 1: Absolute Temperature Conversion Temperature of oxygen in Kelvin: T_O_2 = 273 + 47 = 320 text K$$T_{O_2} = 273 + 47 = 320 \text{ K}$$ ### Step 2: Evaluating for Hydrogen Substitute the given mass ratio and absolute temperature: frac32032 = fracT_H_22$$\frac{320}{32} = \frac{T_{H_2}}{2}$$ 10 = fracT_H_22$$10 = \frac{T_{H_2}}{2}$$ T_H_2 = 20 text K$$T_{H_2} = 20 \text{ K}$$ ### Step 3: Final Conclusion Convert back to Celsius: T_H_2 text in ^circtextC = 20 - 273 = -253^circtextC$$T_{H_2} \text{ in }^{\circ}\text{C} = 20 - 273 = -253^{\circ}\text{C}$$ ### Pattern Recognition Direct proportionality between Temperature and Molecular Mass for equal RMS speeds. T_1/M_1 = T_2/M_2$T_1/M_1 = T_2/M_2$. Always convert to Kelvin before evaluating. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula dP = rho omega^2 x dx, quad PM = rho RT$$dP = \rho \omega^2 x dx, \quad PM = \rho RT$$ ### Core Logic

Setting up differential pressure equation: A[(P+dP) - P] = (dm)(omega^2 x) implies dP = fracdmA omega^2 x$$A[(P+dP) - P] = (dm)(\omega^2 x) \implies dP = \frac{dm}{A} \omega^2 x$$ Substituting rho = fracPMRT$\rho = \frac{PM}{RT}$: int_P_A^P_B fracdPP = fracomega^2 MRT int_0^l x dx$$\int_{P_A}^{P_B} \frac{dP}{P} = \frac{\omega^2 M}{RT} \int_0^l x dx$$ lnleft(fracP_BP_Aright) = fracomega^2 l^2 M2RT implies P_B = P_A e^fracMomega^2 l^22RT$$\ln\left(\frac{P_B}{P_A}\right) = \frac{\omega^2 l^2 M}{2RT} \implies P_B = P_A e^{\frac{M\omega^2 l^2}{2RT}}$$ ### Pattern Recognition Sees: Rotating gas column with centrifugal pressure gradient. Shortcut: Integrate hydrostatic equation with centripetal acceleration term. Check: Matches option (1). ✓ ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula Z = sqrt2pi d^2 N sqrtfrac8RTpi M$$Z = \sqrt{2}\pi d^{2} N \sqrt{\frac{8RT}{\pi M}}$$ where d$d$ is molecular diameter, N$N$ is number density, T$T$ is temperature, and M$M$ is molar mass. ### Core Logic Given that T$T$ and N$N$ are identical for both gases: Z propto fracd^2sqrtM$$Z \propto \frac{d^2}{\sqrt{M}}$$ From the problem statement: d_A = fracd_B2, quad M_A = 4M_B$$d_A = \frac{d_B}{2}, \quad M_A = 4M_B$$ Calculating the ratio of collision frequencies: fracZ_AZ_B = left(fracd_Ad_Bright)^2 times sqrtfracM_BM_A = left(frac12right)^2 times sqrtfrac14 = frac14 times frac12 = frac18$$\frac{Z_A}{Z_B} = \left(\frac{d_A}{d_B}\right)^2 \times \sqrt{\frac{M_B}{M_A}} = \left(\frac{1}{2}\right)^2 \times \sqrt{\frac{1}{4}} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$$ Substituting Z_B = 32 times 10^8 /mathrms$Z_B = 32 \times 10^8 /\mathrm{s}$: Z_A = frac32 times 10^88 = 4 times 10^8 /mathrms$$Z_A = \frac{32 \times 10^8}{8} = 4 \times 10^8 /\mathrm{s}$$ ### Step 1: Final Conclusion The collision frequency in gas A is 4 times 10^8 /mathrms$4 \times 10^8 /\mathrm{s}$. ### Pattern Recognition Proportionality check: Z propto d^2 / sqrtM$Z \propto d^2 / \sqrt{M}$. Diameter halved implies 1/4$\implies 1/4$ factor. Mass quadrupled implies 1/sqrt4 = 1/2$\implies 1/\sqrt{4} = 1/2$ factor. Combined factor = 1/4 times 1/2 = 1/8$= 1/4 \times 1/2 = 1/8$. 32 / 8 = 4$32 / 8 = 4$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula fracP_1 V_1T_1 = fracP_2 V_2T_2$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$ P = P_textatm + rho g h$$P = P_{\text{atm}} + \rho g h$$ ### Core Logic

For an air bubble rising in water, the number of moles of gas remains constant. P_1$P_1$ = Pressure at bottom = P_textatm + rho gh$P_{\text{atm}} + \rho gh$ T_1$T_1$ = Temperature at bottom = 17^circmathrmC = 290 \, mathrmK$17^{\circ}\mathrm{C} = 290 \, \mathrm{K}$ V_1$V_1$ = 2.9 \, mathrmcm^3$2.9 \, \mathrm{cm}^3$ P_2$P_2$ = Pressure at surface = P_textatm$P_{\text{atm}}$ T_2$T_2$ = Temperature at surface = 27^circmathrmC = 300 \, mathrmK$27^{\circ}\mathrm{C} = 300 \, \mathrm{K}$ ### Step 1: Calculate Pressures P_1 = 10^5 + (10^3 times 10 times 5) = 10^5 + 50,000 = 1.5 times 10^5 \, mathrmPa$$P_1 = 10^5 + (10^3 \times 10 \times 5) = 10^5 + 50,000 = 1.5 \times 10^5 \, \mathrm{Pa}$$ P_2 = 10^5 \, mathrmPa$$P_2 = 10^5 \, \mathrm{Pa}$$ ### Step 2: Apply Ideal Gas Law frac(1.5 times 10^5)(2.9)290 = frac(10^5)(V_2)300$$\frac{(1.5 \times 10^5)(2.9)}{290} = \frac{(10^5)(V_2)}{300}$$ frac1.5 times 2.9290 = fracV_2300$$\frac{1.5 \times 2.9}{290} = \frac{V_2}{300}$$ V_2 = frac1.5 times 2.9 times 300290 = 1.5 times 3 = 4.5 \, mathrmcm^3$$V_2 = \frac{1.5 \times 2.9 \times 300}{290} = 1.5 \times 3 = 4.5 \, \mathrm{cm}^3$$ ### Pattern Recognition Standard combined gas law application. Always remember to convert Celsius to Kelvin and carefully compute gauge pressure plus atmospheric pressure for the submerged state. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Mechanical Properties of Fluids

### Related Formula P propto T quad (textGay-Lussac's Law for constant V)$$P \propto T \quad (\text{Gay-Lussac's Law for constant V})$$ ### Core Logic Since the gas is filled in a closed cylinder, the volume V$V$ remains constant. Therefore, P propto T$P \propto T$ (where T must be in Kelvin). Initial temperature T_i = 50^circtextC = 273 + 50 = 323 text K$T_i = 50^{\circ}\text{C} = 273 + 50 = 323 \text{ K}$. Final temperature T_f = 100^circtextC = 273 + 100 = 373 text K$T_f = 100^{\circ}\text{C} = 273 + 100 = 373 \text{ K}$. Note: The phrasing "double its temperature" is slightly ambiguous (Celsius vs Kelvin). As per standard interpretations in such exams, "double its temperature" given in Celsius means 100^circtextC$100^{\circ}\text{C}$ (which is 373 K). If it meant doubling absolute temperature, final would be 646 K. The official interpretation used here doubles the Celsius scale value. ### Step 1: Calculate Final Pressure Using the relation: fracP_fP_i = fracT_fT_i$$\frac{P_f}{P_i} = \frac{T_f}{T_i}$$ fracP_f3.23 times 10^3 = frac373323$$\frac{P_f}{3.23 \times 10^3} = \frac{373}{323}$$ P_f = 3.23 times 10^3 times frac373323 = 3.73 times 10^3 text Pa$$P_f = 3.23 \times 10^3 \times \frac{373}{323} = 3.73 \times 10^3 \text{ Pa}$$ P_f = 3730 text Pa$$P_f = 3730 \text{ Pa}$$ ### Pattern Recognition Be highly alert to 'double temperature' traps when given in Celsius. The question specifically intends 50^circ rightarrow 100^circ$50^\circ \rightarrow 100^\circ$, not 323textK rightarrow 646textK$323\text{K} \rightarrow 646\text{K}$. Calculating with 373K nicely cancels with the 3.23textkPa$3.23\text{kPa}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory Class 11 Physics: Thermodynamics