What is the pressure of 0.540 mol of an ideal gas at 35.5 L and 223 K?

Use P V equals n R T. and R equals 8.314 StartFraction L times kilopascals over moles time K EndFraction..

The pressure of the gas is 0.278 atm

Explanation:

Step 1: Data given

Number of moles of the ideal gas = 0.540 moles

Volume = 35.5 L

Temperature = 223 K

Step 2: Calculate the pressure of the gas

p*V = n*R*T

⇒with p = the pressure of the gas = TO BE DETERMINED

⇒with V= the volume of the gas = 35.5 L

⇒with n = the moles of the gas = 0.540 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature of the gas = 223 K

p = n*R*T / V

p = 0.540 * 0.08206 * 223 / 35.5

p = 0.278 atm

The pressure of the gas is 0.278 atm

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.