At what temperature rms speed of SO3 molecules is 3.16 x 10^2 ms^-1? (R = 8.314 J K^-1 mol^-1)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q127 – AP-EAMCET 2025

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Home Chemistry Kinetic TheoryQuestion #451

At what temperature rms speed of

S O_{3}

molecules is

3.16 \times 10^{2} {ms}^{- 1}

? (

R = 8.314 {J K}^{- 1} {mol}^{- 1}

)

Clue 1 of 3: Kinetic Theory of Gases

The root mean square (RMS) speed of gas molecules is a measure of the average speed of the particles in a gas, directly related to its absolute temperature.

🧠 Full Solution Path

Step 1: Identify Given Values and Target Variable✦ Active

The problem asks for the temperature ( $T$ ) at which $S O_{3}$ molecules have a specific root mean square (RMS) speed. We are given the RMS speed ( $v_{r m s}$ ), the ideal gas constant ( $R$ ), and the gas ( $S O_{3}$ ). We need to calculate the molar mass ( $M$ ) of $S O_{3}$ .

Given values:

v_{r m s} = 3.16 \times 10^{2} {ms}^{- 1}

R = 8.314 {J K}^{- 1} {mol}^{- 1}

💡 Teacher's Secret Hint

Ensure all units are consistent with the SI system before starting calculations. For example, molar mass should be in $kg/mol$ when using $R$ in ${J K}^{- 1} {mol}^{- 1}$ .

Step 2: Calculate Molar Mass of

S O_{3}

○ Expand

First, calculate the molar mass of sulfur trioxide ( $S O_{3}$ ).

Atomic mass of S = $32.06 g/mol$

Atomic mass of O = $16.00 g/mol$

M_{S O_{3}} = (1 \times 32.06) + (3 \times 16.00) = 32.06 + 48.00 = 80.06 g/mol

Convert the molar mass to $kg/mol$ :

M = 80.06 \times 10^{- 3} kg/mol

💡 Teacher's Secret Hint

Remember to convert grams to kilograms for molar mass when using the ideal gas constant $R$ in Joules, as Joules are ${kg m}^{2} s^{- 2}$ .

Step 3: Apply RMS Speed Formula and Rearrange for Temperature○ Expand

The formula for the root mean square speed ( $v_{r m s}$ ) of gas molecules is:

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

To find the temperature ( $T$ ), we need to rearrange this formula. Square both sides:

v_{r m s}^{2} = \frac{3 R T}{M}

Now, solve for $T$ :

T = \frac{M v_{r m s}^{2}}{3 R}

💡 Teacher's Secret Hint

Always double-check your algebraic rearrangement to avoid errors. Squaring both sides is a common first step when the variable is inside a square root.

Step 4: Substitute Values and Calculate Temperature○ Expand

Substitute the calculated molar mass and the given values into the rearranged formula for $T$ :

T = \frac{(80.06 \times 10^{- 3} kg/mol) \times (3.16 \times 10^{2} {ms}^{- 1})^{2}}{3 \times 8.314 {J K}^{- 1} {mol}^{- 1}}

T = \frac{80.06 \times 10^{- 3} \times (3.16)^{2} \times (10^{2})^{2}}{3 \times 8.314}

T = \frac{80.06 \times 10^{- 3} \times 9.9856 \times 10^{4}}{24.942}

T = \frac{80.06 \times 9.9856 \times 10^{1}}{24.942}

T = \frac{7994.9}{24.942}

T \approx 320.54 K

Rounding to a reasonable number of significant figures, the temperature is approximately $320 K$ .

💡 Teacher's Secret Hint

Pay close attention to powers of 10 during calculation. A common mistake is misplacing the decimal or incorrectly handling exponents.

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Chemistry Question 127 – AP-EAMCET 2025

Clue 1 of 3: Kinetic Theory of Gases

🧠 Full Solution Path