Which one of the following graphs is not correct for ideal gas?
d = Density, P = Pressure, T = Temperature

The ideal gas equation can be rewritten to express density by combining the basic gas law with molecular weight properties:

$$PV = nRT = \frac{m}{M}RT$$

Rearranging the equation to solve for density ($$d = \frac{m}{V}$$) gives:

$$P = \frac{m}{V}\frac{RT}{M} \implies P = \frac{dRT}{M}$$

Isolating the density ($$d$$), we get our working formula:

$$d = \left(\frac{M}{R}\right)\frac{P}{T}$$

Step-by-Step Analysis of Each Graph:

• Graph I: $$d$$ vs $$T$$ (at constant $$P$$)

  • From the formula, keeping $$P$$ constant means density is inversely proportional to temperature: $$d \propto \frac{1}{T}$$.
  • The graph of an inverse relationship is a rectangular hyperbola.
  • Status: Graph I is correct.



• Graph II: $$d$$ vs $$T$$ (at constant $$P$$)

  • As established above, $$d$$ and $$T$$ share an inverse relationship ($$d \propto \frac{1}{T}$$). As temperature increases, gas volume expands, causing the density to decrease.
  • Graph II depicts a linear increase with a positive slope ($$d \propto T$$), which is physically impossible for an ideal gas.
  • Status: Graph II is incorrect.



• Graph III: $$d$$ vs $$\frac{1}{T}$$ (at constant $$P$$)

  • Since $$d \propto \frac{1}{T}$$, plotting $$d$$ directly against the inverse of temperature ($$\frac{1}{T}$$) yields a linear relationship ($$y = mx$$).
  • The graph should be a straight line passing through the origin.
  • Status: Graph III is correct.



• Graph IV: $$d$$ vs $$P$$ (at constant $$T$$)

  • Keeping $$T$$ constant means density is directly proportional to pressure: $$d \propto P$$.
  • The graph should be a straight line passing through the origin with a positive slope.
  • Status: Graph IV is correct.

Conclusion:

Because density is inversely related to temperature at a constant pressure, Graph II is geometrically and scientifically flawed.

Answer: Option B — II