Behaviour of Perfect Gases and Kinetic Theory of Gases — Practice Quiz

2 times the original speed

Step 1: vrms = √(3RT/M), so vrms ∝ √T. Step 2: If T becomes 4T, then new vrms = √(3R × 4T / M) = 2 × √(3RT/M). Step 3: So new vrms = 2 × original vrms. Step 4: Option A is wrong (that would apply if vrms ∝ T, not √T). Option C is wrong (4² = 16 is incorrect reasoning). Option D would apply if T doubled, not quadrupled.

(3/2)kT

Step 1: For 1 mole of gas, average kinetic energy = (3/2)RT. Step 2: 1 mole contains N molecules (Avogadro's number). Step 3: Average KE per molecule = (3/2)RT / N = (3/2)(R/N)T = (3/2)kT, where k = R/N is Boltzmann's constant. Step 4: Option A is the KE of 1 mole, not 1 molecule. Option B is missing the factor 3 (that would be for 1 degree of freedom). Option D equals (3/2)kT but written in a confusing form; however, it equals the same thing, yet option C is the standard form.

3

Step 1: Degrees of freedom represent the number of independent ways a molecule can possess energy. Step 2: A monoatomic molecule (single atom) can only translate (move) in 3D space — along X, Y, and Z axes. Step 3: Its rotational kinetic energy is negligible because its moment of inertia is extremely small. Step 4: So it has 3 translational degrees of freedom only. Step 5: 5 degrees of freedom is for diatomic gases, and 6 is for triatomic/polyatomic gases.

(1/2)kT

Step 1: The Law of Equipartition of Energy states that energy is equally distributed among all degrees of freedom. Step 2: The average energy per degree of freedom = (1/2)kT, where k is Boltzmann's constant. Step 3: For a molecule with f degrees of freedom, total average KE = f × (1/2)kT = (f/2)kT. Step 4: For a monoatomic gas (f=3): total KE = (3/2)kT — this is the total, not per degree of freedom. Option A (kT) and D (1/3 kT) are incorrect values.