Exams › NEET › Physics › Kinetic Theory
48 questions with worked solutions.
Answer: R / (γ - 1)
The relationship between Cp, Cv, and γ for an ideal gas is given by Cp - Cv = R and γ = Cp / Cv. Rearranging these equations, Cv = R / (γ - 1).
Answer: 9 / (γ − 1)
The degrees of freedom (f) of a gas molecule are related to the ratio of specific heats (γ) by the formula γ = (f + 2) / f. Rearranging this, f = 2 / (γ - 1). Substituting the options, only D matches correctly.
Q3. The number of translational degrees of freedom for a diatomic gas is
Answer: 3
A diatomic gas has 3 translational degrees of freedom, corresponding to motion along the x, y, and z axes in three-dimensional space.
Q4. What happens to the average velocity of gas molecules as temperature increases?
Answer: It increases.
The average velocity of gas molecules increases with temperature because higher temperature corresponds to greater kinetic energy, which directly affects molecular speed.
Q5. The degree of freedom of a molecule of a triatomic gas is
Answer: 6
A triatomic gas molecule has 6 degrees of freedom: 3 translational and 3 rotational. Linear triatomic molecules lack vibrational modes at room temperature, so only these 6 are considered.
Q6. What is the formula for internal energy (U) of an ideal gas?
Answer: U = (3/2)nRT
For an ideal gas, the internal energy (U) depends only on temperature and is given by U = (3/2)nRT for a monoatomic gas, derived from the degrees of freedom and kinetic theory of gases.
Q7. What happens to the pressure of a gas when its mass is halved and speed is doubled?
Answer: P' = 2P
Pressure is proportional to the product of mass density and the square of the root mean square (rms) speed. Halving the mass reduces density by half, but doubling the speed increases the square of the speed by four times. The net effect is a doubling of pressure.
Q8. What happens to the kinetic energy of molecules at 0 K?
Answer: It becomes zero.
At 0 K (absolute zero), molecular motion ceases, and hence the kinetic energy of the molecules becomes zero as per the kinetic theory of gases.
Q9. What is the escape velocity of oxygen molecules at temperature T?
Answer: V_escape = √(3k_BT/m_0)
Escape velocity is the minimum velocity required for a molecule to escape Earth's gravitational field. The formula √(3k_BT/m_0) relates the root mean square speed of gas molecules to temperature, which is relevant to the context of molecular escape.
Q10. What happens to the rms speed of gas molecules when temperature increases from 300 K to 1200 K?
Answer: It becomes double.
The root mean square (rms) speed of gas molecules is proportional to the square root of the absolute temperature (v_rms ∝ √T). When the temperature increases from 300 K to 1200 K, the ratio of temperatures is 1200/300 = 4. Taking the square root gives √4 = 2, so the rms speed doubles.
Q11. What is the R.M.S. velocity of gas molecules at temperature T2 = 127°C if V1 = 200 m/s at T1 = 27°C?
Answer: V2 = 200 × √(2)
The R.M.S. velocity of gas molecules is proportional to the square root of the absolute temperature. Converting temperatures to Kelvin, T1 = 300 K and T2 = 400 K. Using the relation V2 = V1 × √(T2/T1), we get V2 = 200 × √(400/300) = 200 × √(4/3) = 200 × √(2).
Q12. What is the effect of temperature increase on the kinetic energy of gas molecules?
Answer: It increases.
The kinetic energy of gas molecules is directly proportional to the absolute temperature of the gas. As temperature increases, the average kinetic energy of the molecules also increases.
Q13. In the given (V–T) diagram, what is the relation between pressure P₁ and P₂ ?
Answer: P₂ > P₁
In a V-T diagram (Volume vs. Temperature) for an ideal gas, the slope of the line is inversely proportional to the pressure. If the slope for P₂ is less steep than for P₁, it implies P₂ > P₁.
Answer: 3/4
Using the ideal gas equation, PV = nRT, we know that density (ρ) is given by ρ = PM/RT. Since temperature and R are constant, ρ ∝ PM. Given that the pressure of A is twice that of B (P_A = 2P_B) and the density of A is 1.5 times that of B (ρ_A = 1.5ρ_B), we can write 1.5ρ_B = 2P_B * M_A / RT. Simplifying, M_A / M_B = 3/4.
Answer: 0.2 kg/m³
Using the ideal gas equation, PV = nRT, and substituting n = m/M (where m is mass, M is molar mass), we derive the density formula: ρ = PM/RT. Substituting P = 249000 Pa, M = 2 g/mol = 0.002 kg/mol, R = 8.3 J/mol·K, and T = 300 K, we get ρ = (249000 × 0.002) / (8.3 × 300) = 0.2 kg/m³.
Answer: 1/4 times
Using the ideal gas law, PV = nRT, if the volume is doubled (V → 2V) and the temperature is halved (T → T/2), the pressure P becomes P → P/4. Hence, the pressure becomes 1/4 times its initial value.
Answer: 383/283 x
For an ideal gas, the density divided by pressure is inversely proportional to temperature (in Kelvin). Converting temperatures to Kelvin: 10°C = 283 K and 110°C = 383 K. The ratio at 110°C becomes (283/383) times the original ratio, so the answer is (383/283)x.
Answer: PV = (5/32) RT
The molar mass of oxygen (O2) is 32 g/mol. For 5 g of oxygen, the number of moles (n) is 5/32. Using the ideal gas equation PV = nRT, substituting n = 5/32 gives PV = (5/32)RT.
Q19. At constant volume, temperature is increased then
Answer: number of collisions per unit time will increase
At constant volume, increasing temperature increases the kinetic energy of gas molecules, leading to more frequent collisions with the walls of the container per unit time.
Q20. According to kinetic theory of gases, at absolute zero temperature
Answer: molecular motion stops
According to the kinetic theory of gases, at absolute zero temperature, the kinetic energy of molecules becomes zero, implying that molecular motion ceases completely.