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Reviewed CSCA Physics question · Easy

一定质量的理想气体，初始状态为p₁=1.0×10⁵ Pa，V₁=2.0×10⁻³ m³，T₁=300 K。若气体经历等温变化，压强变为2.0×10⁵ Pa，则体积变为多少？

  1. 1.0×10⁻³ m³
  2. 2.0×10⁻³ m³
  3. 4.0×10⁻³ m³
  4. 0.5×10⁻³ m³
Show the answer and explanation

Correct answer

A. 1.0×10⁻³ m³

Principle or equation

理想气体等温变化：p₁V₁ = p₂V₂（玻意耳定律）。

Why this answer is correct

由p₁V₁ = p₂V₂，得V₂ = p₁V₁/p₂ = (1.0×10⁵ × 2.0×10⁻³) / (2.0×10⁵) = 1.0×10⁻³ m³。

Example

若p₁=2 atm，V₁=3 L，p₂=6 atm，则V₂=1 L。

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