MCQ
From the following $V-T$ diagram we can conclude
  • A
    $P_1 = P_2$
  • $P_1 > P_2$
  • C
    $P_1 < P_2$
  • D
    None of these

Answer

Correct option: B.
$P_1 > P_2$
b
(b) In case of given graph, $V$ and $T$ are related as $V = aT – b$, where $a$ and $b$ are constants.

From ideal gas equation, $PV = \mu RT$

We find $P = \frac{{\mu RT}}{{aT - b}} = \frac{{\mu R}}{{a - b/T}}$

Since $T_2 > T_1$, therefore $P_2 < P_1$.

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