a sample of gas is compressed from 3.25 l to 1.20 l at constant temperature. if the pressure of this gas in the 3.25 l volume is 100.00 kpa, what will the pressure be at 1.20 l ?

You need to re-work this problem. It makes no sense to me as it is. Also, if l is liters, please use a capital L; otherwise, it appears to be another letter.

140kpa

To find the pressure at 1.20 L, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. Mathematically, this can be written as:

P1 × V1 = P2 × V2

Where:
P1 = Initial pressure (100.00 kPa)
V1 = Initial volume (3.25 L)
P2 = Final pressure (to be determined)
V2 = Final volume (1.20 L)

Let's substitute the given values into the equation and solve for P2:

100.00 kPa × 3.25 L = P2 × 1.20 L

Now, solve for P2:

P2 = (100.00 kPa × 3.25 L) / 1.20 L

P2 ≈ 270.83 kPa

Therefore, the pressure of the gas at 1.20 L will be approximately 270.83 kPa.

To determine the pressure of the gas at 1.20 L, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

Mathematically, Boyle's Law can be expressed as:

P1 * V1 = P2 * V2

where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the final pressure and volume.

Given:
P1 = 100.00 kPa (pressure at 3.25 L)
V1 = 3.25 L (initial volume)
V2 = 1.20 L (final volume)

Solving for P2, we rearrange the equation:

P2 = (P1 * V1) / V2

Substituting the values:

P2 = (100.00 kPa * 3.25 L) / 1.20 L

P2 = 271.88 kPa

Therefore, the pressure of the gas at 1.20 L volume will be 271.88 kPa.