Ideal gas change: if absolute temperature is tripled while pressure is simultaneously reduced to one-third, how does the gas volume change?

Introduction / Context:
Quick “ratio” manipulations of the ideal gas law are common in compressor sizing, vessel relief scenarios, and reactor start-up calculations. Recognizing proportionalities avoids algebraic errors and speeds up decision-making under time pressure.

Given Data / Assumptions:

• Ideal gas behavior applies.
• Initial state: P1, V1, T1; final state: P2 = P1/3, T2 = 3*T1.
• Amount of gas n is constant.

Concept / Approach:
For a fixed amount of ideal gas, V ∝ T/P. Therefore, a simultaneous change in T and P scales the volume by the factor (T2/T1) / (P2/P1). Tripling T increases V by a factor of 3; reducing P to one-third increases V by another factor of 3. Combined, the effect is multiplicative, producing a ninefold increase in volume.

Step-by-Step Solution:
Start from PV = nRT → V ∝ T/P (n constant).Compute ratio: (T2/T1) / (P2/P1) = 3 / (1/3) = 9.Thus V2 = 9 * V1.Pick the option indicating ninefold increase.

Verification / Alternative check:
Rearrange: P1V1/T1 = P2V2/T2 → V2 = V1 * (T2/T1) * (P1/P2) = V1 * 3 * 3 = 9 V1, confirming the ratio method.

Why Other Options Are Wrong:
Remain unaltered: ignores both effects.Increase three fold: accounts for only one of the two factors.Decrease three fold: opposite of the true combined effect.

Common Pitfalls:
Using Celsius instead of absolute temperature; adding factors instead of multiplying; changing n inadvertently (e.g., through leaks or phase change).

Final Answer:
Increase nine fold