Which amongst the following options is correct graphical representation of Boyle's Law? (1) P vs T, straight lines through origin labelled V ₁ , V ₂ , V ₃ with V ₁ T ₂ >T ₁ ; (3) P vs 1/V, straight lines labelled T ₃ , T ₂ , T ₁ with T ₃ >T ₂ >T ₁ ; (4) P vs 1/V, hyperbolic curves labelled T ₃ , T ₂ , T ₁ with T ₃ >T ₂ >T ₁ . 1. P vs T lines, V ₁ T ₂ >T ₁ (graph) 3. P vs 1/V lines, T ₃ >T ₂ >T ₁ (graph) 4. P vs 1/V curves, T ₃ >T ₂ >T ₁ (graph)

Answer: 3. P vs 1/V lines, T ₃ >T ₂ >T ₁ (graph) Key idea: Boyle's law: at constant T, P ∝ 1/V , so P plotted against 1/V is a straight line through the origin. The slope is nRT , so a higher temperature gives a steeper line: T₃ > T₂ > T₁ . This matches option 3. Option 1 shows P vs T at constant V, which is Gay-Lussac's law, not Boyle's. Option 2 is a hyperbola for P vs V, but the labels are reversed: the curve farthest from the axes should be the highest temperature, yet it is labelled T₁ . Option 4 shows P vs 1/V as curves, but P is directly proportional to 1/V, so it must be a straight line.

Topic: Chemical equations and stoichiometry· Some Basic Concepts in Chemistry

Which amongst the following options is correct graphical representation of Boyle's Law?(1) P vs T, straight lines through origin labelled V_1, V_2, V_3 with V_1<V_2<V_3; (2) P vs V, hyperbolas labelled T_1, T_2, T_3 with T_3>T_2>T_1; (3) P vs 1/V, st…

Which amongst the following options is correct graphical representation of Boyle's Law?

(1) P vs T, straight lines through origin labelled V\(_1\), V\(_2\), V\(_3\) with V\(_1\)<V\(_2\)<V\(_3\); (2) P vs V, hyperbolas labelled T\(_1\), T\(_2\), T\(_3\) with T\(_3\)>T\(_2\)>T\(_1\); (3) P vs 1/V, straight lines labelled T\(_3\), T\(_2\), T\(_1\) with T\(_3\)>T\(_2\)>T\(_1\); (4) P vs 1/V, hyperbolic curves labelled T\(_3\), T\(_2\), T\(_1\) with T\(_3\)>T\(_2\)>T\(_1\).

four graphs (options 1-4)

1. P vs T lines, V\(_1\)<V\(_2\)<V\(_3\) (graph)

2. P vs V curves, T\(_3\)>T\(_2\)>T\(_1\) (graph)

3. P vs 1/V lines, T\(_3\)>T\(_2\)>T\(_1\) (graph)

4. P vs 1/V curves, T\(_3\)>T\(_2\)>T\(_1\) (graph)

Doubtly Asked 30 Oct 2026 NEET UG 2023 neet Save WhatsApp Report

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Answer: 3. P vs 1/V lines, T\(_3\)>T\(_2\)>T\(_1\) (graph)

Key idea: Boyle's law: at constant T, \(P \propto \frac1V\), so P plotted against 1/V is a straight line through the origin.

The slope is \(nRT\), so a higher temperature gives a steeper line: \(T_3 > T_2 > T_1\). This matches option 3.

Option 1 shows P vs T at constant V, which is Gay-Lussac's law, not Boyle's.
Option 2 is a hyperbola for P vs V, but the labels are reversed: the curve farthest from the axes should be the highest temperature, yet it is labelled \(T_1\).
Option 4 shows P vs 1/V as curves, but P is directly proportional to 1/V, so it must be a straight line.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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Which amongst the following options is correct graphical representation of Boyle's Law?(1) P vs T, straight lines through origin labelled V_1, V_2, V_3 with V_1<V_2<V_3; (2) P vs V, hyperbolas labelled T_1, T_2, T_3 with T_3>T_2>T_1; (3) P vs 1/V, st…

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