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Calculate the temperature at which the root mean square veloci

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Solving time: 5 mins

Calculate the temperature at which the root mean square velocity of $SO_{3}$ gas molecules becomes equal to the average velocity of argon gas molecules at $2 7^{\circ} C$ . $(M_{SO_{3}} = 80 g mol^{- 1}; M_{Ar} = 40 g mol^{- 1})$

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Calculate the temperature at which the root mean square velocity of

SO_{3}

gas molecules becomes equal to the average velocity of argon gas molecules at

2 7^{\circ} C

(M_{SO_{3}} = 80 g mol^{- 1}; M_{Ar} = 40 g mol^{- 1})

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

Solution
RMS velocity of

SO_{3}

gas molecules

= \frac{3 RT}{M} = \frac{3 RT}{80}

Average velocity of "Ar" gas molecules

= \frac{8 R \times 300}{\frac{22}{7} \times 40}

\Rightarrow \frac{3 RT}{80} T = \frac{8 R \times 300}{\frac{22}{7} \times 40} = \frac{8 \times 300 \times 7 \times 80}{22 \times 40 \times 3} = 509.1 K

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Question Text	Calculate the temperature at which the root mean square velocity of $SO_{3}$ gas molecules becomes equal to the average velocity of argon gas molecules at $2 7^{\circ} C$ . $(M_{SO_{3}} = 80 g mol^{- 1}; M_{Ar} = 40 g mol^{- 1})$
Topic	States of Matter
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1
Upvotes	106

Calculate the temperature at which the root mean square velocity of $SO_{3}$ gas molecules becomes equal to the average velocity of argon gas molecules at $2 7^{\circ} C$ . $(M_{SO_{3}} = 80 g mol^{- 1}; M_{Ar} = 40 g mol^{- 1})$

Text solutionVerified

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Calculate the temperature at which the root mean square velocity of $SO_{3}$ gas molecules becomes equal to the average velocity of argon gas molecules at $2 7^{\circ} C$ . $(M_{SO_{3}} = 80 g mol^{- 1}; M_{Ar} = 40 g mol^{- 1})$