Chemistry Q126 – AP-EAMCET 2026

Home Chemistry Kinetic TheoryQuestion #450

The rms velocity of

S O_{2}

molecules at

400 K

is same as the most probable velocity of

O_{2}

molecules at

T (K)

. What is the kinetic energy (in kJ) of

5

moles of

O_{2}

T (K)

? (At.wt:

S = 32 u

O = 16 u

), (

R = 8.3 {J K}^{- 1} {mol}^{- 1}

)

Clue 1 of 3: Gas Velocities and Kinetic Energy

Recall the formulas for root mean square velocity ( $v_{r m s}$ ), most probable velocity ( $v_{m p}$ ), and the total kinetic energy for an ideal gas.

🧠 Full Solution Path

Step 1: Calculate Molar Masses✦ Active

First, we need to calculate the molar masses ( $M$ ) of $S O_{2}$ and $O_{2}$ from their given atomic weights. Atomic weight of $S = 32 u$ and $O = 16 u$ .

M_{S O_{2}} = 32 + (2 \times 16) = 32 + 32 = 64 g/mol = 0.064 kg/mol

M_{O_{2}} = 2 \times 16 = 32 g/mol = 0.032 kg/mol

💡 Teacher's Secret Hint

Remember to convert molar mass from $g/mol$ to $kg/mol$ when using the ideal gas constant $R = 8.3 {J K}^{- 1} {mol}^{- 1}$ in formulas for molecular velocities to ensure unit consistency for joules (which is ${kg m}^{2} s^{- 2}$ ). However, in ratio equations, units often cancel.

Step 2: Determine Temperature T(K) for O2○ Expand

The problem states that the rms velocity of $S O_{2}$ at $400 K$ is equal to the most probable velocity of $O_{2}$ at $T (K)$ . We use the respective formulas for these velocities:

v_{r m s} = \sqrt{\frac{3 R T}{M}}

v_{m p} = \sqrt{\frac{2 R T}{M}}

Equating them for the given conditions:

\sqrt{\frac{3 R \times 400}{M_{S O_{2}}}} = \sqrt{\frac{2 R \times T}{M_{O_{2}}}}

Squaring both sides and canceling $R$ :

\frac{3 \times 400}{M_{S O_{2}}} = \frac{2 \times T}{M_{O_{2}}}

Substitute the molar masses in $g/mol$ (they will cancel out consistently):

\frac{3 \times 400}{64} = \frac{2 \times T}{32}

\frac{1200}{64} = \frac{T}{16}

18.75 = \frac{T}{16}

T = 18.75 \times 16 = 300 K

💡 Teacher's Secret Hint

In equations where molar mass appears in both the numerator and denominator or when comparing ratios, using molar mass in $g/mol$ is often fine as the units will cancel. However, always be cautious and convert to SI units ( $kg/mol$ ) if absolute values for velocity or energy are needed.

Step 3: Calculate Kinetic Energy of O2○ Expand

Now, we need to find the kinetic energy of $5$ moles of $O_{2}$ at the calculated temperature $T = 300 K$ . The total translational kinetic energy for $n$ moles of an ideal gas is given by:

K E = n \frac{3}{2} R T

Given $n = 5 mol$ , $R = 8.3 {J K}^{- 1} {mol}^{- 1}$ , and $T = 300 K$ :

K E = 5 \times \frac{3}{2} \times 8.3 {J K}^{- 1} {mol}^{- 1} \times 300 K

K E = 5 \times 1.5 \times 8.3 \times 300

K E = 7.5 \times 8.3 \times 300

K E = 18675 J

💡 Teacher's Secret Hint

The constant $R$ is given in ${J K}^{- 1} {mol}^{- 1}$ , so the calculated kinetic energy will be in Joules (J). The question asks for the kinetic energy in kiloJoules (kJ).

Step 4: Convert Kinetic Energy to kJ○ Expand

Finally, convert the kinetic energy from Joules to kiloJoules:

K E (kJ) = \frac{18675 J}{1000 J/kJ} = 18.675 kJ

💡 Teacher's Secret Hint

Pay close attention to the units requested in the final answer (e.g., kJ vs J, atm vs Pa) to avoid common conversion errors.

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Chemistry Question 126 – AP-EAMCET 2026

Clue 1 of 3: Gas Velocities and Kinetic Energy

🧠 Full Solution Path