Reviewed CSCA Physics question · Easy

A rigid container holds 1.5 mol of an ideal gas at a pressure of 2.0 × 10^5 Pa and a temperature of 300 K. The gas is heated at constant volume until its pressure becomes 3.0 × 10^5 Pa. What is the new temperature of the gas?

200 K
450 K
600 K
900 K

Show the answer and explanation

Correct answer

B. 450 K

Principle or equation

For an ideal gas at constant volume, pressure is directly proportional to absolute temperature: P1/T1 = P2/T2.

Why this answer is correct

Given P1 = 2.0×10^5 Pa, T1 = 300 K, P2 = 3.0×10^5 Pa. Using P1/T1 = P2/T2 gives T2 = T1 × (P2/P1) = 300 K × (3.0/2.0) = 450 K.

Example

If a gas at 200 K and 1 atm is heated at constant volume until its pressure doubles, the new temperature is 400 K.

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，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。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。