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 From the Ideal Gas Law equation: P V = n R T implies P = fracn R TV$$P V = n R T \implies P = \frac{n R T}{V}$$ ### Core Logic Given that both vessels possess the same volume (V_A = V_B$V_A = V_B$) and identical temperature states (T_A = T_B$T_A = T_B$), the ratio of pressure reduces directly to: fracP_AP_B = fracn_An_B$$\frac{P_A}{P_B} = \frac{n_A}{n_B}$$ where n_A$n_A$ and n_B$n_B$ are the number of moles of Hydrogen and Oxygen respectively. ### Step 1: Calculate the Number of Moles For Hydrogen (H_2$H_2$, molar mass = 2 mathrm~g/mol$2 \mathrm{~g/mol}$): n_A = frac12$$n_A = \frac{1}{2}$$ For Oxygen (O_2$O_2$, molar mass = 32 mathrm~g/mol$32 \mathrm{~g/mol}$): n_B = frac132$$n_B = \frac{1}{32}$$ ### Step 2: Find the Pressure Ratio Substituting mole counts into the direct ratio: fracP_AP_B = frac1/21/32 = frac322 = 16$$\frac{P_A}{P_B} = \frac{1/2}{1/32} = \frac{32}{2} = 16$$ Therefore, the pressure ratio fracP_AP_B$\frac{P_A}{P_B}$ is 16$16$. ### Pattern Recognition For gas mixtures or vessel comparisons under constant volume and temperature, pressure matches the molar abundance directly (P propto n$P \propto n$). Remember that standard elementary gases (H_2, O_2, N_2$H_2, O_2, N_2$) exist as diatomic configurations when defining molar values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula f_textmixture = fracn_1 f_1 + n_2 f_2n_1 + n_2$$f_{\text{mixture}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$$ gamma_textmixture = 1 + frac2f_textmixture$$\gamma_{\text{mixture}} = 1 + \frac{2}{f_{\text{mixture}}}$$ ### Core Logic For a monoatomic gas, degrees of freedom f_1 = 3$f_1 = 3$. Moles n_1 = 3$n_1 = 3$. For a diatomic gas, degrees of freedom f_2 = 5$f_2 = 5$. Moles n_2 = 2$n_2 = 2$. We can compute the equivalent degrees of freedom for the mixture using a weighted average. ### Step 1: Calculate Equivalent Degrees of Freedom f_textmixture = fracn_1 f_1 + n_2 f_2n_1 + n_2$$f_{\text{mixture}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$$ f_textmixture = frac3(3) + 2(5)3 + 2 = frac9 + 105 = frac195$$f_{\text{mixture}} = \frac{3(3) + 2(5)}{3 + 2} = \frac{9 + 10}{5} = \frac{19}{5}$$ ### Step 2: Calculate Adiabatic Exponent gamma_textmixture = 1 + frac2f_textmixture$$\gamma_{\text{mixture}} = 1 + \frac{2}{f_{\text{mixture}}}$$ gamma_textmixture = 1 + frac2frac195 = 1 + frac1019 = frac2919 approx 1.52$$\gamma_{\text{mixture}} = 1 + \frac{2}{\frac{19}{5}} = 1 + \frac{10}{19} = \frac{29}{19} \approx 1.52$$ ### Pattern Recognition Alternatively, you can compute C_v$C_v$ and C_p$C_p$ for the mixture: C_v,mix = fracn_1 C_v1 + n_2 C_v2n_1 + n_2$C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$, and gamma_mix = fracC_p,mixC_v,mix$\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}}$. Both methods yield identical results rapidly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

### Related Formula v_textrms = sqrtfrac3RTM$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$ ### Core Logic For the RMS velocities to be equal, the ratio of temperature to molar mass (T/M$T/M$) must be identical for both gases. ### Step 1: Set Up Equivalency sqrtfrac3RT_H_2M_H_2 = sqrtfrac3RT_O_2M_O_2$$\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$$ fracT_H_2M_H_2 = fracT_O_2M_O_2$$\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$$ ### Step 2: Substitute Values T_O_2 = 47^circmathrmC = 47 + 273 = 320 mathrm~K$T_{O_2} = 47^\circ\mathrm{C} = 47 + 273 = 320 \mathrm{~K}$ M_H_2 = 2 mathrm~g/mol$M_{H_2} = 2 \mathrm{~g/mol}$ M_O_2 = 32 mathrm~g/mol$M_{O_2} = 32 \mathrm{~g/mol}$ fracT_H_22 = frac32032$$\frac{T_{H_2}}{2} = \frac{320}{32}$$ T_H_2 = 2 times 10 = 20 mathrm~K$$T_{H_2} = 2 \times 10 = 20 \mathrm{~K}$$ ### Pattern Recognition v_textrms$v_{\text{rms}}$ scales strictly as sqrtT/M$\sqrt{T/M}$. Remember to always convert Celsius to Kelvin before substituting. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

### Related Formula U = n C_v T$U = n C_v T$ where C_v = fracf2 R$C_v = \frac{f}{2} R$ and f$f$ is the degree of freedom. ### Core Logic Argon (Ar) is monatomic implies f_1 = 3 implies C_v_1 = frac3R2$\implies f_1 = 3 \implies C_{v_1} = \frac{3R}{2}$. Oxygen (O_2$O_2$) is diatomic implies f_2 = 5$\implies f_2 = 5$ (neglecting vibrational modes) implies C_v_2 = frac5R2$\implies C_{v_2} = \frac{5R}{2}$. Total internal energy U = U_1 + U_2 = n_1 C_v_1 T + n_2 C_v_2 T$U = U_1 + U_2 = n_1 C_{v_1} T + n_2 C_{v_2} T$. ### Step 1: Compute Total Energy U = 8 times left(frac3R2right) T + 6 times left(frac5R2right) T$$U = 8 \times \left(\frac{3R}{2}\right) T + 6 \times \left(\frac{5R}{2}\right) T$$ U = 4(3RT) + 3(5RT)$$U = 4(3RT) + 3(5RT)$$ U = 12RT + 15RT$U = 12RT + 15RT$ U = 27 RT$U = 27 RT$ ### Pattern Recognition Internal energy is strictly additive. Immediately map Monatomic to 3/2$\to 3/2$ and Diatomic to 5/2$\to 5/2$. Plug linearly: 8(1.5) + 6(2.5) = 12 + 15 = 27$8(1.5) + 6(2.5) = 12 + 15 = 27$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

### Related Formula textKE = fracf2kT$$\text{KE} = \frac{f}{2}kT$$ ### Core Logic The average translational kinetic energy of any gas molecule depends only on the absolute temperature of the gas and is independent of the nature or mass of the gas. For 1 mole of any ideal gas, the average translational kinetic energy is frac32RT$\frac{3}{2}RT$. Therefore, at a given temperature, the kinetic energy parameter is uniform across all ideal gases. ### Pattern Recognition Temperature is directly proportional to average translational kinetic energy. If T is constant, KE is constant for all gases regardless of mass. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory