gamma_A is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. gamma_B is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If fracgamma_Agamma_B = left(1 + frac1nright) then the value of n is

Numerical Answer Type:
Enter a numerical value Answer: 3 to 3 +4 marks

Solution & Explanation

### Related Formula gamma = 1 + frac2f ### 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 (translational only) gamma_A = 1 + frac23 = frac53 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 degrees of freedom. - Therefore, the total active degrees of freedom is: f_B = 3 + 3 + 2 = 8 The specific heat ratio of B is: gamma_B = 1 + frac2f_B = 1 + frac28 = 1 + frac14 = frac54 Now, find the ratio of specific heat capacities: fracgamma_Agamma_B = frac5/35/4 = frac43 We are given: fracgamma_Agamma_B = 1 + frac1n implies frac43 = 1 + frac1n implies frac1n = frac13 implies n = 3 ### Step 1: Final Conclusion The value of 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). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

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Kinetic Theory Of Gases

More Kinetic Theory of Gases Previous-Year Questions — Page 4

Q50 jee_main_2024_29_jan_morning Ideal Gas Equation
Two vessels A and B are of the same size and are at same temperature. A contains 1 mathrm~g of hydrogen and B contains 1 mathrm~g of oxygen. P_A and P_B are the pressures of the gases in A and B respectively, then fracP_AP_B is:
  • A. 16
  • B. 8
  • C. 4
  • D. 32

Solution

### Related Formula From the Ideal Gas Law equation: P V = n R T implies P = fracn R TV ### Core Logic Given that both vessels possess the same volume (V_A = V_B) and identical temperature states (T_A = T_B), the ratio of pressure reduces directly to: fracP_AP_B = fracn_An_B where n_A and n_B are the number of moles of Hydrogen and Oxygen respectively. ### Step 1: Calculate the Number of Moles For Hydrogen (H_2, molar mass = 2 mathrm~g/mol): n_A = frac12 For Oxygen (O_2, molar mass = 32 mathrm~g/mol): n_B = frac132 ### Step 2: Find the Pressure Ratio Substituting mole counts into the direct ratio: fracP_AP_B = frac1/21/32 = frac322 = 16 Therefore, the pressure ratio fracP_AP_B is 16. ### Pattern Recognition For gas mixtures or vessel comparisons under constant volume and temperature, pressure matches the molar abundance directly (P propto n). Remember that standard elementary gases (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
Q46 jee_main_2024_30_january_evening Mixture of Gases
If three moles of monoatomic gas left(gamma = frac53right) is mixed with two moles of a diatomic gas left(gamma = frac75right), the value of adiabatic exponent gamma for the mixture is:
  • A. 1.75
  • B. 1.40
  • C. 1.52
  • D. 1.35

Solution

### Related Formula f_textmixture = fracn_1 f_1 + n_2 f_2n_1 + n_2 gamma_textmixture = 1 + frac2f_textmixture ### Core Logic For a monoatomic gas, degrees of freedom f_1 = 3. Moles n_1 = 3. For a diatomic gas, degrees of freedom f_2 = 5. Moles 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_textmixture = frac3(3) + 2(5)3 + 2 = frac9 + 105 = frac195 ### Step 2: Calculate Adiabatic Exponent gamma_textmixture = 1 + frac2f_textmixture gamma_textmixture = 1 + frac2frac195 = 1 + frac1019 = frac2919 approx 1.52 ### Pattern Recognition Alternatively, you can compute C_v and C_p for the mixture: C_v,mix = fracn_1 C_v1 + n_2 C_v2n_1 + n_2, and gamma_mix = fracC_p,mixC_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
Q49 jee_main_2024_30_jan_morning RMS Velocity of Gases
At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at 47^circmathrmC?
  • A. 80 mathrm~K
  • B. -73 mathrm~K
  • C. 4 mathrm~K
  • D. 20 mathrm~K

Solution

### Related Formula v_textrms = sqrtfrac3RTM ### Core Logic For the RMS velocities to be equal, the ratio of temperature to molar mass (T/M) must be identical for both gases. ### Step 1: Set Up Equivalency sqrtfrac3RT_H_2M_H_2 = sqrtfrac3RT_O_2M_O_2 fracT_H_2M_H_2 = fracT_O_2M_O_2 ### Step 2: Substitute Values T_O_2 = 47^circmathrmC = 47 + 273 = 320 mathrm~K M_H_2 = 2 mathrm~g/mol M_O_2 = 32 mathrm~g/mol fracT_H_22 = frac32032 T_H_2 = 2 times 10 = 20 mathrm~K ### Pattern Recognition v_textrms scales strictly as sqrtT/M. Remember to always convert Celsius to Kelvin before substituting. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases
Q38 jee_main_2024_31_jan_evening Internal Energy of a Gas Mixture
A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is
  • A. 29 text RT
  • B. 20 text RT
  • C. 27 text RT
  • D. 21 text RT

Solution

### Related Formula U = n C_v T where C_v = fracf2 R and f is the degree of freedom. ### Core Logic Argon (Ar) is monatomic implies f_1 = 3 implies C_v_1 = frac3R2. Oxygen (O_2) is diatomic implies f_2 = 5 (neglecting vibrational modes) implies C_v_2 = frac5R2. Total internal energy 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 = 4(3RT) + 3(5RT) U = 12RT + 15RT U = 27 RT ### Pattern Recognition Internal energy is strictly additive. Immediately map Monatomic to 3/2 and Diatomic to 5/2. Plug linearly: 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
Q31 jee_main_2024_31_jan_morning Kinetic Energy Of Gas Molecules
The parameter that remains the same for molecules of all gases at a given temperature is :
  • A. textkinetic energy
  • B. textmomentum
  • C. textmass
  • D. textspeed

Solution

### Related Formula textKE = fracf2kT ### 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. 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

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