Skip to content
NCERT Exemplar · Q5

Q.For a fixed mass of an ideal gas, its volume VV (vertical axis, in litres) is plotted against its absolute temperature TT (horizontal axis, in kelvin) at two different constant pressures P1P_1 and P2P_2. Both plots are straight lines passing through the origin. The line labelled P2P_2 is steeper (has the larger slope) than the line labelled P1P_1. What can be inferred about the relation between P1P_1 and P2P_2?

(a) P1>P2P_1 > P_2
(b) P1=P2P_1 = P_2
(c) P1<P2P_1 < P_2
(d) data is insufficient.
Yanam BieapMCQ· 1mImportance★★★★★est
48% · 24/50 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

On a VV versus TT plot at constant pressure, the ideal-gas law gives a straight line through the origin whose slope is nR/PnR/P. The larger the slope, the smaller the pressure. Since the P2P_2 line is the steeper one, P2P_2 is the lower pressure, so P1>P2P_1 > P_2.

Concept

For a fixed amount of ideal gas at constant pressure, volume is directly proportional to absolute temperature (Charles' law), which plots as a straight line through the origin.

Why this formula

From PV=nRTPV = nRT at constant PP:

V=(nRP)T,V = \left(\frac{nR}{P}\right) T,

so the slope of the VV–TT line is

slope=nRP.\text{slope} = \frac{nR}{P}.

With nn and RR fixed, slope is inversely proportional to PP: a steeper line corresponds to a smaller pressure.

Steps

  1. Both lines pass through the origin, confirming V∝TV \propto T for each fixed pressure.
  2. Slope =nR/P⇒P=nR/slope= nR/P \Rightarrow P = nR/\text{slope}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.