$I, I I$ and $I I I$ are three isobars repectively at $p_{1} . p_{2}$ and $p_{3}$ as shown. Arrange $p_{1}, p_{2}$ and $p_{3}$ in increasing order.

p_{1} < p_{2} < p_{3}

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p_{3} < p_{2} < p_{1}

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p_{1} = p_{2} = p_{3}

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p_{3} < p_{1} < p_{2}

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Solution

The correct option is B $p_{3} < p_{2} < p_{1}$
Solution:- (B) $p_{3} < p_{2} < p_{1}$
From ideal gas equation,
$P v = n R T$
For same volume,
$P \propto \frac{1}{T}$
As from the graph-
$T_{I I I} > T_{I I} > T_{I}$
Thus $p_{1} > p_{2} > p_{3}$

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I,II and III are three isobars repectively at p1.p2 and p3 as shown. Arrange p1,p2 and p3 in increasing order.

The correct option is B p3<p2<p1Solution:- (B) p3<p2<p1From ideal gas equation,Pv=nRTFor same volume,P∝1TAs from the graph-TIII>TII>TIThus p1>p2>p3

$I, I I$ and $I I I$ are three isobars repectively at $p_{1} . p_{2}$ and $p_{3}$ as shown. Arrange $p_{1}, p_{2}$ and $p_{3}$ in increasing order.

The correct option is B $p_{3} < p_{2} < p_{1}$
Solution:- (B) $p_{3} < p_{2} < p_{1}$
From ideal gas equation,
$P v = n R T$
For same volume,
$P \propto \frac{1}{T}$
As from the graph-
$T_{I I I} > T_{I I} > T_{I}$
Thus $p_{1} > p_{2} > p_{3}$