The given figure represents two isobaric processes for the same mass of an ideal gas, then

Solution

Physics sir · 4 steps

On a V–T graph the slope is nR/P, so the steeper line belongs to the lower pressure.

1. Read the slope. For a fixed mass of ideal gas, PV = nRT gives V = (nR/P)T. At constant pressure this is a straight line through the origin, and its slope is nR/P.

   slope = V/T = nR/P ⇒ P ∝ 1/slope

2. Compare the lines. The mass is the same, so nR is the same for both lines. The P₂ line is steeper than the P₁ line, so P₂ has the larger slope and therefore the smaller pressure.
3. Check with one temperature. Pick any temperature T and draw a vertical line. The P₂ line gives the larger volume. Same gas, same T, larger V means lower pressure.
4. So. P₁ > P₂, which is option (b).

The idea to keep. Before comparing lines on any gas graph, write the law with the plotted quantities as y and x. Whatever sits in the slope tells you the order, and here pressure sits in the denominator.

Answer from the NTA final answer key. Solution by PrepWiser.