For a particular ideal gas which of the following graphs represents the variation of mean squared velocity of the gas…

Q: For a particular ideal gas which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?

### Related Formula ### Core Logic The parameter asked is the mean squared velocity, which corresponds directly to Vrms2. From the ideal gas kinematics relation, we observe: Comparing this format against standard geometric linear templates (y = mx), the curve must map as a clean straight line originating from absolute zero zero coordinates. ### Step 1: Final Conclusion This linear profile matches Graph (1), selecting option (1). ### Pattern Recognition Watch the vertical ordinate labels carefull

Q5 jee_main_2025_28_jan_evening Translational Kinetic Energy

The kinetic energy of translation of the molecules in

50 ~g

of

CO₂

gas at

17° C

is :

A. $3986.3 J$ $3986.3 \mathrm{J}$
B. $4102.8 J$ $4102.8 \mathrm{J}$
C. $4205.5 ~J$ $4205.5 \mathrm{~J}$
D. $3582.7 J$ $3582.7 \mathrm{J}$

Solution

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.translational = (3)/(2) n R T

Core Logic

Given data:

Mass of $CO₂$ $\mathrm{CO}_2$ gas, $m = 50 g$ $m = 50 \text{ g}$
Molar mass of $CO₂$ $\mathrm{CO}_2$, $M = 44 g/mol$ $M = 44 \text{ g/mol}$
Absolute Temperature, $T = 17 + 273.15 = 290.15 K$ $T = 17 + 273.15 = 290.15 \text{ K}$
Universal gas constant, $R \approx 8.314 J/(mol )$

Calculate total moles $n$ $n$:

n = (50)/(44) \approx 1.1364 moles

Substitute values into the expression :

K.E.translational = (3)/(2) \times (50)/(44) \times 8.314 \times 290.15

K.E.translational = 1.5 \times 1.1364 \times 8.314 \times 290.15 \approx 4108.6 J

The closest matching value specified in the test alternatives is $4102.8 J$ $4102.8 \mathrm{J}$.

Pattern Recognition

A common mistake is using $(5)/(2)nRT$ $\frac{5}{2}nRT$ or $(7)/(2)nRT$ $\frac{7}{2}nRT$ because $CO₂$ $\mathrm{CO}_2$ is a triatomic linear molecule. Remember that translational kinetic energy is always $(3)/(2)nRT$ $\frac{3}{2}nRT$ for any gas sample, as translation has exactly 3 degrees of freedom regardless of the molecular layout.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

Q jee_main_2025_29_jan_morning Ideal Gas Laws

A container of fixed volume contains a gas at

27°C

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

°C

Numerical Answer. Answer: 327 to 327

Solution

Related Formula

(P₁)/(T₁) = (P₂)/(T₂)

Core Logic

Initial temperature $T₁ = 27 + 273 = 300 K$ $T_1 = 27 + 273 = 300\text{ K}$. Since volume is kept fixed :

(P)/(300) = (2P)/(T₂) T₂ = 600 K

Converting back to Celsius :

T₂ = 600 - 273 = 327°C

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. $(9)/(4) R$ $\frac{9}{4} R$
B. $(7)/(4) R$ $\frac{7}{4} R$
C. $(3)/(2) R$ $\frac{3}{2} R$
D. $(5)/(2) R$ $\frac{5}{2} R$

Solution

Related Formula

Molar specific heat at constant volume for a gas mixture:

CV, mix = n₁ CV1 + n₂ CV2n₁ + n₂

For a monoatomic gas:

CV1 = (3)/(2)R

For a diatomic gas:

CV2 = (5)/(2)R

Core Logic

Given values: $n₁ = 2$ $n_1 = 2$ (monoatomic), $n₂ = 6$ $n_2 = 6$ (diatomic).

Substitute these inputs directly into the mixture equation:

CV, mix = (2 \times ((3)/(2)R) + 6 \times ((5)/(2)R))/(2 + 6)

Step 1: Simplify Expression

CV, mix = (3R + 15R)/(8) = (18R)/(8) = (9)/(4)R

Pattern Recognition

Weighted average rule based on internal degrees of freedom: total internal energy changes scale additively with mole numbers.

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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 10²⁵

molecules per cubic meter at

1.38 atm

(Given,

k = 1.38 \times 10⁻²³ J K⁻¹

) is:

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

Solution

Related Formula

The state equation of an ideal gas in terms of the number of molecules $N$ $N$ and Boltzmann constant $k$ $k$ is:

$PV = NkT$ $PV = NkT$

Rearranging to express pressure in terms of number density $n = N/V$ $n = N/V$:

$P = n k T$ $P = n k T$

Core Logic

Given parameters:

Number density, $n = (N)/(V) = 2.0 \times 10²⁵ molecules/m³$
Pressure, $P = 1.38 atm = 1.38 \times 1.01 \times 10⁵ N/m²$
Boltzmann constant, $k = 1.38 \times 10⁻²³ J K⁻¹$

Step 1: Solve for Temperature

Rearranging $P = n k T$ $P = n k T$ for temperature $T$ $T$:

T = (P)/(nk)

Substitute the values:

T = 1.38 \times 1.01 \times 10⁵(2.0 \times 10²⁵) \times (1.38 \times 10⁻²³)

Notice that the term $1.38$ $1.38$ cancels out from numerator and denominator:

T = (1.01 \times 10⁵)/(2.0 \times 10²)

T = (1.01 \times 10³)/(2.0) \approx (1010)/(2) \approx 500 K

Pattern Recognition

The numerical values are designed to cancel out smoothly. Spotting the $1.38$ $1.38$ cancelation instantly saves valuable calculation time.

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 $feq$ $f_{\text{eq}}$ for a mixture of gases is:

feq = (n₁ f₁ + n₂ f₂)/(n₁ + n₂)

where:

$n₁, n₂$ $n_1, n_2$ are the number of moles of each gas.
$f₁, f₂$ $f_1, f_2$ are the respective degrees of freedom of each gas.

Core Logic

For the given gases:

Polyatomic gas:
Moles, $n₁ = N$ $n_1 = N$
Degrees of freedom, $f₁ = 6$ $f_1 = 6$
Monoatomic gas:
Moles, $n₂ = 2$ $n_2 = 2$
Degrees of freedom, $f₂ = 3$ $f_2 = 3$

We want the mixture to behave as a diatomic gas. For a diatomic gas:

Equivalent degrees of freedom, $feq = 5$ $f_{\text{eq}} = 5$

Step 1: Solve for N

Substitute the values into the degrees of freedom mixture formula:

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

5(N + 2) = 6N + 6

5N + 10 = 6N + 6

10 - 6 = 6N - 5N 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.

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases

For a particular ideal gas which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

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More Kinetic Theory Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Pattern Recognition

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Solution

Related Formula

Core Logic

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Solution

Related Formula

Core Logic

Step 1: Simplify Expression

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Solve for Temperature

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Solve for N

Pattern Recognition

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