Question

) In addition to ongoing sources, such as car engines, pollutants can also be introduced

into the air through isolated incidents like the rupture of a gas storage tank. Most
gases are stored and transported at high pressures. A common laboratory cylinder of
methane, for example, has a volume of 49.0 L and is filled to a pressure of 154 atm.
Suppose that all the CH4 from this cylinder is released and expands until its pressure
falls to 1.00 atm. What volume would the CH4 occupy?

Answer 1

Using Boyle's law, the final volume of methane (CH4) released from a cylinder at 154 atm and 49.0 L to 1.00 atm pressure would be 7556 L.

Step-by-step explanation:

To find out what volume the methane (CH4) would occupy at 1.00 atm after being released from a storage tank where it was initially at a volume of 49.0 L and a pressure of 154 atm, we can use the ideal gas law and the concept of Boyle's law. Boyle's law states that for a given mass of gas at constant temperature, the volume of the gas is inversely proportional to the pressure. This means P1 * V1 = P2 * V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume respectively.

Applying these principles:

• Initial Pressure (P1) = 154 atm
• Initial Volume (V1) = 49.0 L
• Final Pressure (P2) = 1.00 atm
• Final Volume (V2) = Unknown

To find the final volume V2, rearrange the equation:

V2 = (P1 * V1) / P2

V2 = (154 atm * 49.0 L) / 1.00 atm

V2 = 7556 L

This is the volume occupied by the methane gas after the expansion to 1.00 atm pressure.

Answer 2

V₂ = 7546 L

Step-by-step explanation:

Given data:

Initial volume = 49.0 L

Initial pressure = 154 atm

Final pressure = 1 atm

Final volume = ?

Solution:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

154 atm × 49.0 L = 1.00 atm × V₂

V₂ = 7546 atm. L/1 atm

V₂ = 7546 L