This problem can be solved using Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship when temperature is held constant.

Boyle's Law is represented by the formula P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Given in the problem:
P1 = 1.20 x 10^2 kPa
V1 = 1.50 x 10^3 mL
P2 = 1.00 x 10^2 kPa

We are asked to solve for V2.

First, convert V1 from mL to L because the answer should be in L. 1 L = 1000 mL, so V1 = 1.50 x 10^3 mL / 1000 = 1.50 L.

Now, substitute the known values into Boyle's Law and solve for V2:

(1.20 x 10^2 kPa)(1.50 L) = (1.00 x 10^2 kPa)(V2)

Solving for V2 gives:

V2 = (1.20 x 10^2 kPa)(1.50 L) / 1.00 x 10^2 kPa = 1.80 L

So, the volume of the gas at a pressure of 1.00 x 10^2 kPa is 1.80 L.

Question

This problem can be solved using Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship when temperature is held constant.

Boyle's Law is represented by the formula P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Given in the problem:
P1 = 1.20 x 10^2 kPa
V1 = 1.50 x 10^3 mL
P2 = 1.00 x 10^2 kPa

We are asked to solve for V2.

First, convert V1 from mL to L because the answer should be in L. 1 L = 1000 mL, so V1 = 1.50 x 10^3 mL / 1000 = 1.50 L.

Now, substitute the known values into Boyle's Law and solve for V2:

(1.20 x 10^2 kPa)(1.50 L) = (1.00 x 10^2 kPa)(V2)

Solving for V2 gives:

V2 = (1.20 x 10^2 kPa)(1.50 L) / 1.00 x 10^2 kPa = 1.80 L

So, the volume of the gas at a pressure of 1.00 x 10^2 kPa is 1.80 L.

Knowee AI · Accepted Answer

This problem can be solved using Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship when temperature is held constant.

Boyle's Law is represented by the formula P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Given in the problem:
P1 = 1.20 x 10^2 kPa
V1 = 1.50 x 10^3 mL
P2 = 1.00 x 10^2 kPa

We are asked to solve for V2.

First, convert V1 from mL to L because the answer should be in L. 1 L = 1000 mL, so V1 = 1.50 x 10^3 mL / 1000 = 1.50 L.

Now, substitute the known values into Boyle's Law and solve for V2:

(1.20 x 10^2 kPa)(1.50 L) = (1.00 x 10^2 kPa)(V2)

Solving for V2 gives:

V2 = (1.20 x 10^2 kPa)(1.50 L) / 1.00 x 10^2 kPa = 1.80 L

So, the volume of the gas at a pressure of 1.00 x 10^2 kPa is 1.80 L.

Chemistry I14) Calculate the volume of a gas (in L) at a pressure of 1.00 x 10^2 kPa if its volume at 1.20 x 10^2 kPa is 1.50 x 10^3 mL. 1.80 L 1.90 L 1.85 L 1.75 LPrevious Question Next QuestionDo not use the 'Forward' or 'Back' button on your browser.

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