Oxygen weighing 10 g, at a temperature of 272 K and a pressure of 100 kPa, is compressed isothermally

Oxygen weighing 10 g, at a temperature of 272 K and a pressure of 100 kPa, is compressed isothermally to a volume of 1.2 liters. Determine the oxygen pressure after compression. Consider oxygen as a diatomic ideal gas with a molar mass of 32g.

m = 10 g = 0.01 kg.

T1 = 272 ° C.

P1 = 100 kPa = 100,000 Pa.

V2 = 1.2 L = 0.0012 m ^ 3.

R = 8.31 m ^ 2 * kg / s ^ 2 * ° K * mol.

M = 0.032 kg / mol.

P2 -?

With isothermal compression, the Boyle-Tariott gas law is valid: P1 * V1 = P2 * V2.

P2 = P1 * V1 / V2.

Let’s write down the Mendeleev-Cliperon equation: P1 * V1 = m * R * T1 / M and express the initial volume from it: V1 = m * R * T1 / M * P1.

P2 = P1 * m * R * T1 / V2 * M * P1 = m * R * T1 / V2 * M.

P2 = 0.01 kg * 8.31 m ^ 2 * kg / s ^ 2 * ° K * mol * 272 ° K / 0.0012 m ^ 3 * 0.032 kg / mol = 588625 Pa = 588 kPa.

Answer: after compression, the oxygen pressure became P2 = 588625 Pa.