JEE Main 2022 · 27 June · evening · Single correct · +4 / −1
For a perfect gas, two pressures P₁ and P₂ are shown in figure. The graph shows :
- (a) P₁ > P₂
- (b) P₁ < P₂
- (c) P₁ = P₂
- (d) Insufficient data to draw any conclusion
Answer and solution
Answer (a) P₁ > P₂
Solution
Physics sir · 3 steps
On a V–T graph the slope of an isobar is nR/P, so the steeper line belongs to the lower pressure.
- Slope of an isobar. For a fixed amount of ideal gas, PV = nRT gives V = nR/PT. At constant pressure, V against T is a straight line.
slope = V/T = nR/P ⇒ slope ∝ 1/P
- Read the graph. The P₂ line is steeper than the P₁ line, so 1/P₂ > 1/P₁. Another check: at the same temperature, the P₂ line shows a larger volume, and a larger volume at the same T means a lower pressure.
- So. P₂ < P₁, that is, P₁ > P₂. The answer is (a).
The idea to keep. On V–T plots the slope is nR/P, and on P–T plots the slope is nR/V. In both, the steeper line has the smaller value of the quantity held constant.
Answer from the NTA final answer key. Solution by PrepWiser.