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Standard XII

Physics

Pressure vs Velocity

3.0 g of an i...

Question

$3.0 g$ of an ideal gas $A$ , when placed in an evacuated vessel at $27^{\circ} C$ exerts a pressure of $1.5 atm$ . When $9.0 g$ of another ideal gas $B$ , non- reacting with $A$ , is introduced in the same vessel pressure becomes $2.0 atm$ . What is the ratio between the average speeds of $A$ and $B$ at the same temperature?

1 : 3

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1 : 4

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4 : 1

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3 : 1

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Solution

The correct option is D $3 : 1$
Let, pressure of gas $A$ be $P_{A}$ and pressure of gas $B$ be $P_{B}$ .
So,
$P_{A} V = \frac{3}{M_{A}} R T P_{B} V = \frac{9}{M_{B}} R T But P_{B} = 2.0 - 1.5 = 0.5 atm S o, \frac{P_{A}}{P_{B}} = \frac{1.5}{0.5} = \frac{3 M_{B}}{9 M_{A}} \Rightarrow \frac{M_{B}}{M_{A}} = 9 Therefore, average speed, \frac{V_{A}}{V_{B}} = \sqrt{\frac{M_{B}}{M_{A}}} = 3$

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3.0 g of an ideal gas A, when placed in an evacuated vessel at 27 ∘C exerts a pressure of 1.5 atm. When 9.0 g of another ideal gas B, non- reacting with A, is introduced in the same vessel pressure becomes 2.0 atm. What is the ratio between the average speeds of A and B at the same temperature?

The correct option is D 3:1Let, pressure of gas A be PA and pressure of gas B be PB. So, PAV=3MARTPBV=9MBRTBut PB=2.0−1.5=0.5 atmSo, PAPB=1.50.5=3MB9MA⇒MBMA=9Therefore, average speed, VAVB=√MBMA=3