Reviewed CSCA Physics question · Standard

一定质量的理想气体，初始状态为 p1 = 2×10⁵ Pa，V1 = 3 L，T1 = 300 K。若气体经历等温变化，压强变为 p2 = 6×10⁵ Pa，求末态体积 V2。

Show the answer and explanation

Correct answer

A. 1 L

Principle or equation

玻意耳定律（等温变化）：p1V1 = p2V2。

Why this answer is correct

由 p1V1 = p2V2 得 V2 = p1V1 / p2 = (2×10⁵ × 3) / (6×10⁵) = 6/6 = 1 L。

Example

若压强变为4×10⁵ Pa，则 V2 = 1.5 L。

This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.

Related practice questions

Standard · Thermal physicsA sealed container holds 2 moles of an ideal gas at a pressure of 1.0 × 10⁵ Pa and a temperature of…Standard · Thermal physicsAn ideal gas occupies 2.0 L at a pressure of 1.0 atm and a temperature of 300 K. If the amount of…Easy · Thermal physics一定质量的理想气体，在温度不变的情况下，体积增大为原来的2倍，则压强变为原来的多少倍？Standard · Thermal physics一定质量的理想气体，在压强不变的情况下，温度从27°C升高到127°C，求体积变为原来的多少倍。Standard · Thermal physics一定质量的理想气体，在压强不变的情况下，温度从27°C升高到127°C。求体积变为原来的多少倍。