How many grams of krypton gas will occupy 2.55 liters at 1.25 atm if it occupies 3.75 liters at 0.750 atm at 255 K.

An actual solution would be appreciated rather than the glib comment to the question being asked.

94.9 grams of krypton gas will occupy 2.55 liters at 1.25 atm

Explanation:

According to Boyle - Mariotte's law, for a constant temperature system, the pressure changed inversely proportional to the volume. By decreasing the volume of the container, the pressure increases because the molecules collide more against the wall of the container

This is the Ideal Gas law equation.

P. V = n. R. T

In the first case, the gas occupes 0,13 moles

3,75L. 0,750 atm = n. 0,082L. atm/mol. K. 255K

(3,75L. 0,750 atm) / 0,082mol. K/L. atm. 255K = n = 0,13 moles

The new moles that the gas occupes will be higher than 0,13 so according to Boyle-Mariotte's law (at the same T°, we suppose that)

P1. V1 = P2. V2

P1. V1 = n1. R. T

P2. V2 = n2. R. T

(P1. V1) / R. T = (P2. V2) / n2. R. T

R and T are the same on both sides so we cancel them.

(P1. V1) = (P2. V2) / n2

n2 = (P2. V2) / (P1. V1)

n2 (2,55. 1,25) / (3,75. 0,750)

n2 = 1,13 moles

Molar mass of Kr = 83,80 g/m

Molar mass. moles = gr - --> 83,80 g/m. 1,13 moles = 94,9 g