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JEE Main 20127 MayOnline
Exam
JEE Main
Year
2012
Date
7 May
Mode
Online
Subject
Chemistry
Chapter
States of Matter

JEE Main 2012 · 7 May · Online · Chemistry · States of Matter

For 1 ~mol of an ideal gas at a constant temperature T , the plot...

For 1 mol1 \mathrm{~mol} of an ideal gas at a constant temperature TT, the plot of (log⁡P)(\log P) against (log⁡V)(\log V) is a ( PP : Pressure, VV : Volume)

Options

  1. AStraight line parallel to xx-axis.
  2. BStraight line with a negative slope.
  3. CCurve starting at origin.
  4. DStraight line passing through origin.

Correct answer

B. Straight line with a negative slope.

Step-by-step solution

According to Boyle's law, PV=P V= constant ∴log⁡P+log⁡V=\therefore \log P+\log V= constant log⁡P=−log⁡V+\log P=-\log V+ constant Hence, the plot of log⁡P\log P vs log⁡V\log V is straight line with negative slope.
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