Verified Solution • 2025–26 Session moderate

QJEEM22Chc16c: JEE Main 2022 Chemistry - Class 11 JEEM22Chc16c Solution: Which amongst the given plots is the correct plot for pressure…

Which amongst the given plots is the correct plot for pressure (p) vs density (d) for an ideal gas?

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/bfd5c548c927427191b639122096f181d0058bd3be74afb29360aab311253140.webp
  • B https://ik.imagekit.io/novatra/photos/4514e925c6115d7013cf53256bb86807f58e03786a1cd3f08815066ff59f00a6.webp
  • C https://ik.imagekit.io/novatra/photos/199bd8970fcec1fa11790a8b016925eb33bacb082cdec96e23bb1bdf0ffd7e1d.webp
  • D https://ik.imagekit.io/novatra/photos/3b16212a2527139d50d65229f658d34e12be05c692fb4fc77b0214ad145fa456.webp

Step-by-Step Verified Solution

From ideal gas equation we know,

PV = nRT

⇒PV=WMRT\Rightarrow PV = {W \over M}RT

⇒P=WV . RTM\Rightarrow P = {W \over V}\,.\,{{RT} \over M}

⇒P=d . RTM\Rightarrow P = d\,.\,{{RT} \over M} [ ∵\because d=WVd = {W \over V} ]

For a fixed amount of gas at a fixed temperature, M and T is constant.

∴\therefore P∝dP \propto d

So graph between pressure (P) and density (d) is a straight line.

Also, Pd=RTM{P \over d} = {{RT} \over M}

⇒Pd∝T\Rightarrow {P \over d} \propto T [as R, M = constant]

∴\therefore When temperature T increases then slope of graph between P and d increases.

As T3 > T2 > T1 then slope of T3 will be highest then T2 and then T1.

So, graph (B) is the right graph.

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