gammaA is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. gammaB is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If then the value of n is

Let's find the specific heat ratio for each gas based on degrees of freedom: 1. **Monoatomic gas A:** - Degrees of freedom, fA = 3 (translational only) 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 x 1 = 2 degrees of freedom. - Therefore, the total active degrees of freedom is: The specific heat ratio of B is: Now, find the ratio of specific heat capacities: We are given: The value of n is 3. Always remember that each vibrational mode contributes exactly 2 degrees of freedom because it holds both kinetic and potential energy components (fvib = 2 x modes). Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

gammaA is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. gammaB is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of f

Let's find the specific heat ratio for each gas based on degrees of freedom: 1. **Monoatomic gas A:** - Degrees of freedom, fA = 3 (translational only) 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 x 1 = 2 degrees of freedom. - Therefore, the total active degrees of freedom is: The specific heat ratio o

gammaA is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. gammaB is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If then the value of n is

Let's find the specific heat ratio for each gas based on degrees of freedom: 1. **Monoatomic gas A:** - Degrees of freedom, fA = 3 (translational only) 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 x 1 = 2 degrees of freedom. - Therefore, the total active degrees of freedom is: The specific heat ratio of B is: Now, find the ratio of specific heat capacities: We are given: The value of n is 3. Always remember that each vibrational mode contributes exactly 2 degrees of freedom because it holds both kinetic and potential energy components (fvib = 2 x modes). Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

gammaA is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. gammaB is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of f

Let's find the specific heat ratio for each gas based on degrees of freedom: 1. **Monoatomic gas A:** - Degrees of freedom, fA = 3 (translational only) 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 x 1 = 2 degrees of freedom. - Therefore, the total active degrees of freedom is: The specific heat ratio o

gammaA is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. gammaB is the specific…

Q jee_main_2025_29_jan_morning Ideal Gas Laws

A container of fixed volume contains a gas at

27^circmathrmC

. To double the pressure of the gas, the temperature of gas should be raised to _________

^circmathrmC

Numerical Answer. Answer: 327 to 327

Solution

### Related Formula

fracP_1T_1 = fracP_2T_2

### Core Logic Initial temperature

T_1 = 27 + 273 = 300text K

. Since volume is kept fixed :

fracP300 = frac2PT_2 implies T_2 = 600text K

Converting back to Celsius :

T_2 = 600 - 273 = 327^circmathrmC

### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

Q36 jee_main_2024_01_february_morning Specific Heat Capacity

Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is:

A. $frac94 R$ $\frac{9}{4} R$
B. $frac74 R$ $\frac{7}{4} R$
C. $frac32 R$ $\frac{3}{2} R$
D. $frac52 R$ $\frac{5}{2} R$

Solution

### Related Formula Molar specific heat at constant volume for a gas mixture:

C_Vtext, mix = fracn_1 C_V1 + n_2 C_V2n_1 + n_2

For a monoatomic gas:

C_V1 = frac32R

For a diatomic gas:

C_V2 = frac52R

### Core Logic Given values:

n_1 = 2

(monoatomic),

n_2 = 6

(diatomic). Substitute these inputs directly into the mixture equation:

C_Vtext, mix = frac2 times left(frac32Rright) + 6 times left(frac52Rright)2 + 6

### Step 1: Simplify Expression

C_Vtext, mix = frac3R + 15R8 = frac18R8 = frac94R

### Pattern Recognition Weighted average rule based on internal degrees of freedom: total internal energy changes scale additively with mole numbers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

Q35 jee_main_2024_29_january_evening Ideal Gas Equation and Temperature

The temperature of a gas having

2.0 times 10^25

molecules per cubic meter at

1.38text atm

(Given,

k = 1.38 times 10^-23text J K^-1

) is:

A. $500text K$ $500\text{ K}$
B. $200text K$ $200\text{ K}$
C. $100text K$ $100\text{ K}$
D. $300text K$ $300\text{ K}$

Solution

### Related Formula The state equation of an ideal gas in terms of the number of molecules

N

and Boltzmann constant

k

is:

PV = NkT

Rearranging to express pressure in terms of number density

n = N/V

:

P = n k T

### Core Logic Given parameters: * Number density,

n = fracNV = 2.0 times 10^25text molecules/m^3

* Pressure,

P = 1.38text atm = 1.38 times 1.01 times 10^5text N/m^2

* Boltzmann constant,

k = 1.38 times 10^-23text J K^-1

### Step 1: Solve for Temperature Rearranging

P = n k T

for temperature

T

:

T = fracPnk

Substitute the values:

T = frac1.38 times 1.01 times 10^5(2.0 times 10^25) times (1.38 times 10^-23)

Notice that the term

1.38

cancels out from numerator and denominator:

T = frac1.01 times 10^52.0 times 10^2

T = frac1.01 times 10^32.0 approx frac10102 approx 500text K

### Pattern Recognition The numerical values are designed to cancel out smoothly. Spotting the

1.38

cancelation instantly saves valuable calculation time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

Q48 jee_main_2024_29_january_evening Degrees of Freedom and Specific Heat of Gas Mixtures

N

moles of a polyatomic gas (

f = 6

) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of

N

is:

A. $6$ $6$
B. $3$ $3$
C. $4$ $4$
D. $2$ $2$

Solution

### Related Formula The equivalent degrees of freedom

f_texteq

for a mixture of gases is:

f_texteq = fracn_1 f_1 + n_2 f_2n_1 + n_2

where: *

n_1, n_2

are the number of moles of each gas. *

f_1, f_2

are the respective degrees of freedom of each gas. ### Core Logic For the given gases: 1. Polyatomic gas: * Moles,

n_1 = N

* Degrees of freedom,

f_1 = 6

2. Monoatomic gas: * Moles,

n_2 = 2

* Degrees of freedom,

f_2 = 3

We want the mixture to behave as a diatomic gas. For a diatomic gas: * Equivalent degrees of freedom,

f_texteq = 5

### Step 1: Solve for N Substitute the values into the degrees of freedom mixture formula:

5 = frac(N)(6) + (2)(3)N + 2

5(N + 2) = 6N + 6

5N + 10 = 6N + 6

10 - 6 = 6N - 5N implies N = 4

### Pattern Recognition Diatomic equivalent degree of freedom is 5. Since the monoatomic degrees of freedom (3) and polyatomic degrees of freedom (6) bracket 5, you can use the weighted ratio method to find the molar proportions directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

Q50 jee_main_2024_27_jan_morning Kinetic Energy and Temperature

The average kinetic energy of a monatomic molecule is

0.414text eV

at temperature:

A. $3000text K$ $3000\text{ K}$
B. $3200text K$ $3200\text{ K}$
C. $1600text K$ $1600\text{ K}$
D. $1500text K$ $1500\text{ K}$

Solution

### Related Formula

K_textavg = frac32 k_B T

### Core Logic Given energy is in electron-volts (

1text eV = 1.6 times 10^-19text J

), we isolate

T

:

T = frac2 K_textavg3 k_B

Substitute constants (

k_B = 1.38 times 10^-23text J/K

): ### Step 1: Compute value

T = frac2 times 0.414 times 1.6 times 10^-193 times 1.38 times 10^-23

T = frac1.3248 times 10^-194.14 times 10^-23 = 0.32 times 10^4 = 3200text K

### Pattern Recognition Converting eV energy properties straight to structural SI standard Joules reveals highly cleanly simplified scalar components when paired with Boltzmann values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases

Solution & Explanation

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