Closed, rigid container heating:\nA gas initially at 27°C in a sealed rigid vessel is heated to 300°C. How does the pressure change (assume ideal gas)?

Introduction / Context:
In constant-volume heating, ideal-gas pressure scales directly with absolute temperature. This is fundamental in sizing pressure relief and checking vessel ratings during heat-up transients.

Given Data / Assumptions:

• Closed, rigid container (constant volume, constant moles).
• Ideal-gas approximation.
• T1 = 27°C = 300 K; T2 = 300°C = 573 K.

Concept / Approach:
For ideal gases at constant V and n, P ∝ T (absolute). Hence, P2/P1 = T2/T1. Compute the ratio using Kelvin temperatures.

Step-by-Step Solution:
Convert to Kelvin: T1 = 273 + 27 = 300 K; T2 = 273 + 300 = 573 K.Compute P2/P1 = 573/300 ≈ 1.91.This is very close to 2; among discrete options, “doubled” is the correct selection.

Verification / Alternative check:
Using PV = nRT at constant V and n shows pressure changes linearly with T, confirming the ratio above.

Why Other Options Are Wrong:
“Halved” and “trebled” contradict P ∝ T; “unpredictable” is incorrect because n and V are fixed; “increased by 50%” underestimates the effect.

Common Pitfalls:
Using Celsius directly instead of Kelvin; forgetting the vessel is rigid (constant V).

Final Answer:
doubled