One mole of an ideal diatomic gas is taken through the cycle as shown in the figure- 1 - 2- isochoric process 2 - 3- a straight line of P-V diagram 3 - 1- isobaric processThe average molecular speed of the gas in the states 1-2 and 3 in the ratio-

The correct option is A 1:2:2 State 1Let temperature be #160; T_o State 2Since 1 2 is isochoric process #160; P/T=Constant Thus #160; T ...

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Standard XII

Physics

RMS and KE Expressions

One mole of a...

Question

One mole of an ideal diatomic gas is taken through the cycle as shown in the figure:

$1 \to 2 :$ isochoric process
$2 \to 3 :$ a straight line of P-V diagram
$3 \to 1 :$ isobaric process

The average molecular speed of the gas in the states 1,2 and 3 in the ratio:

1 : 2 : 2

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1 : \sqrt{2} : \sqrt{2}

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1 : 1 : 1

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1 : 2 : 4

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Solution

The correct option is A $1 : 2 : 2$
State 1
Let temperature be $T_{o}$
State 2
Since 1-2 is isochoric process $\frac{P}{T} = C o n s t a n t$
Thus $T e m p e r a t u r e = 4 T_{o}$

State 3
Since 3-1 is an isobaric process $\frac{V}{T} = C o n s t a n t$
Thus $T e m p e r a t u r e = 4 T_{o}$
The average molecular speed is directly proportional to $\sqrt{\frac{R T}{M}}$
$\Rightarrow v \propto \sqrt{T}$
Hence
$S t a t e 1 v \propto T_{o}$
$S t a t e 2 v \propto \sqrt{4 T_{o}} = 2 \sqrt{T_{o}}$
$S t a t e 3 v \propto \sqrt{4 T_{o}} = 2 \sqrt{T_{o}}$
$\Rightarrow R a t i o = 1 : 2 : 2$
Thus (a) is the correct answer.

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One mole of an ideal diatomic gas is taken through the cycle as shown in the figure:1→2: isochoric process2→3: a straight line of P-V diagram3→1: isobaric processThe average molecular speed of the gas in the states 1,2 and 3 in the ratio:

One mole of an ideal diatomic gas is taken through the cycle as shown in the figure:

$1 \to 2 :$ isochoric process
$2 \to 3 :$ a straight line of P-V diagram
$3 \to 1 :$ isobaric process

The average molecular speed of the gas in the states 1,2 and 3 in the ratio: