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Question: Answered & Verified by Expert
Certain amount of an ideal gas is taken from its initial state 1 to final state 4 through the paths $1 \rightarrow 2 \rightarrow 3 \rightarrow 4$ as shown in figure. $A B, C D, E F$ are all isotherms. If $v_p$ is the most probable speed of the molecules, then
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Options:
  • A $v_p$ at $3=v_p$ at $4>v_p$ at $2>v_p$ at 1
  • B $v_p$ at $3>v_p$ at $1>v_p$ at $2>v_p$ at 4
  • C $v_p$ at $3>v_p$ at $2>v_p$ at $4>v_p$ at 1
  • D $v_p$ at $2=v_p$ at $3>v_p$ at $1>v_p$ at 4
Solution:
2685 Upvotes Verified Answer
The correct answer is: $v_p$ at $3=v_p$ at $4>v_p$ at $2>v_p$ at 1
$v_p \propto \sqrt{T}$

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