Which one of the following changes would cause the pressure of a gas to double assuming temperature was held constant? A) Decreasing the volume of a gas from 4.0 L to 2.0 L while keeping moles the same. B) Increasing the volume of a gas from 2.0 L to 4.0 L while keeping moles the same. C) Doubling the number of moles of gas present while decreasing the volume from 2.0 L to 1.0 L. D) Doubling both the moles of gas and the volume of gas.

Statement A is correct. Decreasing the volume of a gas from 4.0 L to 2.0 L while keeping moles the same.

Explanation:

Step 1: Data given

Pressure of a gas is doubled

Temperature stays constant

Step 2:

P1*V1 / n1 = P2*V2/n2

⇒with P1 = the initial pressure = 1.0 atm

⇒with V1 = the initial volume

⇒with n1 = the initial number of moles gas

⇒with P2 = the final pressure = 2.0 atm

⇒with V2 = the final volume

⇒with n2 = the final number of moles

Decreasing the volume of a gas from 4.0 L to 2.0 L while keeping moles the same.

Suppose the initial number of moles = 2.0 moles

1.0 * 4.0 / 2.0 = 2.0 *2.0 / n2

The final number of moles n2 = 2.0 moles = the same

This statement is true

Increasing the volume of a gas from 2.0 L to 4.0 L while keeping moles the same.

Suppose the initial number of moles = 2.0 moles

1.0 * 2.0 / 2.0 = 2.0 *4.0 / n2

The final number of moles n2 = 8.0 moles = not the same

This statement is false

C) Doubling the number of moles of gas present while decreasing the volume from 2.0 L to 1.0

Suppose the initial number of moles = 2.0 moles

1.0 * 2.0 / 2.0 = 2.0 *1.0 / n2

The final number of moles n2 = 2.0 moles = the same

This statement is false

D) Doubling both the moles of gas and the volume of gas.

Suppose the initial number of moles = 2.0 moles

The initial volume = 1.0 L

1.0 * 1.0 / 2.0 = 2.0 *2.0 / n2

The final number of moles n2 = 8.0 moles = not the same

This statement is false

Statement A is correct. Decreasing the volume of a gas from 4.0 L to 2.0 L while keeping moles the same.

there all right good

explanation:

58750.08068 is the answer i think