Question

A sample of helium has has a volume of 4.0 L at a pressure of 4.0 atm. The temperature and the number of gas particles do not change. What is the volume of gas (in L) at 5.8 atm? Round to 2 significant figures

Answer 1

Using Boyle's Law, the final volume of helium at 5.8 atm is found to be approximately 2.76 L, rounded to two significant figures.

Step-by-step explanation:

To determine the new volume of a gas when the pressure changes, we can use Boyle's Law, which states that the volume of a gas varies inversely with its pressure when temperature and amount of gas particles remain constant. The formula for Boyle's Law is P1V1 = P2V2, where P represents pressure, V represents volume, and the subscripts 1 and 2 represent the initial and final states, respectively.

In the given problem, a sample of helium with an initial volume (V1) of 4.0 L and initial pressure (P1) of 4.0 atm undergoes a pressure increase to 5.8 atm (P2). Since the temperature and number of gas particles remain constant, we can rearrange the Boyle's Law equation to solve for the final volume (V2) using the steps below:

1. Write the equation P1V1 = P2V2.
2. Plug in the given values: (4.0 atm)(4.0 L) = (5.8 atm)(V2).
3. Solve for V2: V2 = (4.0 atm)(4.0 L) / 5.8 atm.
4. Calculate the result: V2 = 16 L / 5.8 atm ≈ 2.76 L.

Therefore, the final volume of the helium gas at 5.8 atm pressure will be approximately 2.76 L, rounded to two significant figures.