Ideal gas law: Sometimes an experiment requires a certain pure gas to be used at reduced pressure. One way to achieve this is to purchase a sealed glass container filled with the gas, and to introduce the gas into a vacuum by attaching the glass container to the vacuum chamber and breaking the tip of the glass container using a metallic bean and a magnet. If the volume of the glass container is 1.0 L and it is at a pressure of 1.0 × 10⁵ Pa and if the vacuum chamber has a volume of 2.0 L, what will the pressure be after the gas, which is to be assumed to be an ideal gas, is released into the vacuum chamber and the temperature has returned to its original value? (Note that the glass container remains part of the system.) A) 33 kPa B) 50 kPa C) 300 kPa D) 200 kPa

Question

Quizplus · Accepted Answer

According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. Since the temperature and number of moles of gas remain constant, we can write:P1V1 = P2V2where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.Substituting the given values, we get:(1.0 × 10^5 Pa)(1.0 L) = P2(3.0 L)Solving for P2, we get:P2 = 33 kPaTherefore, the correct answer is A) 33 kPa.

Ideal Gas Law: Sometimes an Experiment Requires a Certain Pure