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

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