Ideal-gas change: If a gas's absolute temperature is doubled while its pressure is simultaneously reduced to one-half, how does its volume change?

Introduction / Context:
Combined gas law questions test the ability to relate pressure, volume, and temperature for an ideal gas. Engineers constantly use these relations for compressor sizing, storage calculations, and reactor volume changes. Recognizing proportionalities quickly can save time and avoid algebraic errors.

Given Data / Assumptions:

• Ideal gas behavior applies.
• Initial state: P1, V1, T1. Final state: P2 = P1/2, T2 = 2*T1.
• Gas amount n is constant.

Concept / Approach:
The ideal gas law rearranged gives V ∝ T/P for fixed n. Apply the specified scalings to find the multiplicative effect on volume. The linearity of volume with respect to T and inverse proportionality with respect to P allows a quick ratio method without solving for absolute values.

Step-by-Step Solution:
Start with V ∝ T/P for fixed n.Compute factor: (T2/T1) / (P2/P1) = (2) / (1/2) = 4.Therefore, V2 = 4 * V1.Select the option indicating a fourfold increase.

Verification / Alternative check:
Plug into PV = nRT twice: P1V1/T1 = P2V2/T2 → V2 = V1 * (T2/T1) * (P1/P2) = V1 * 2 * 2 = 4 V1, confirming the ratio reasoning.

Why Other Options Are Wrong:
No change or doubling: underestimate the combined effect of increased T and decreased P.One-fourth of original: inverts the correct proportionality.

Common Pitfalls:
Mixing up absolute and relative temperature (always use absolute temperature); applying Celsius differences directly to proportional relations; forgetting that halving pressure increases volume for a fixed amount of gas.

Final Answer:
Fourfold increase