A sample of methane gas CH4, occupies a volume (L) of 58.2 L at a pressure of 7.25 atm. What volume will the gas occupy if the pressure is lowered to 2.03 atm

To solve this problem, you can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

Boyle's Law can be represented by the equation:

P1 * V1 = P2 * V2

Where:
P1 and V1 are the initial pressure and volume respectively,
P2 and V2 are the final pressure and volume respectively.

Let's plug in the values given in the problem:

P1 = 7.25 atm
V1 = 58.2 L
P2 = 2.03 atm
V2 = ?

Now, rearrange the formula to solve for V2:

V2 = (P1 * V1) / P2

Substitute the known values into the equation:

V2 = (7.25 atm * 58.2 L) / 2.03 atm

Calculate the result:

V2 = 208.05 L / 2.03 atm

V2 ≈ 102.42 L

Therefore, if the pressure is lowered to 2.03 atm, the volume of the methane gas CH4 will be approximately 102.42 L.

To solve this problem, we can use Boyle's law, which states that the pressure and the volume of a gas are inversely proportional at constant temperature. The equation can be written as:

P1 * V1 = P2 * V2

where:
P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume

In this case, we know:
P1 = 7.25 atm
V1 = 58.2 L
P2 = 2.03 atm
V2 = unknown

Using Boyle's law, we can solve for V2:

P1 * V1 = P2 * V2
7.25 atm * 58.2 L = 2.03 atm * V2

Dividing both sides of the equation by 2.03 atm:

V2 = (7.25 atm * 58.2 L) / 2.03 atm

V2 = 209.84 L

Therefore, the volume of the gas will be 209.84 L when the pressure is lowered to 2.03 atm.

P1V1 = P2V2