25°C, 1.00 mole of O2 gas is found occupy a volume of 12.5 L at a pressure of 198K PA. What is the value of the gas constant in L times K PA divided by more times K

8.305 L.KPa/mol.K

Explanation:

Step 1:

Data obtained from the question. This includes:

Temperature (T) = 25°C

Number of mole (n) = 1 mole

Volume (V) = 12.5 L

Pressure (P) = 198 KPa

Gas constant (R) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

K = °C + 273

K = 25°C + 273

K = 298K

Therefore, the temperature is 298K

Step 3:

Determination of the gas constant.

Applying the ideal gas equation PV = nRT, the gas constant can be obtained as follow:

PV = nRT

198KPa x 12.5L = 1mol x R x 298K

Divide both side 1mol x 298K

R = (198KPa x 12.5L) / (1mol x 298K)

R = 8.305L.KPa/mol.K

Therefore the gas constant is 8.305L.KPa/mol.K

.

The gas constant has a value of 0.082 L*atm/mol*K

Explanation:

Step 1: Data given

Temperature = 25.0 °C = 298.15 K

Number of moles O2 gas = 1.00 moles

Volume = 12.5 L

Pressure = 198 Kpa = 1.95411 atm

Step 2: Calculate the gas constant

p*V = n*R*T

⇒with p = the pressure of the gast = 1.95411 atm

⇒with V = the volume of the gas = 12.5 L

⇒with n = the number of moles of O2 gas = 1.00 moles

⇒with R = the gas constant = TO BE DETERMINED

⇒with T = the temperature =  25 °C = 298.15 K

R = (p*V) / (n*T)

R = (1.95411 * 12.5) / (1.00 * 298.15)

R = 0.082 L*atm*mol*K

The gas constant has a value of 0.082 L*atm/mol*K

samuel de champlain. in 1608, king henry iv sponsored pierre dugua, sieur de mons and samuel de champlain as founders of the city of quebec with 28 men. this was the second permanent french settlement in the colony of canada.