Ideal Gas Law
Ideal Gas Law: The ideal gas law is the equation of state of a hypothetical ideal gas. It obeys Boyle's Law and Charles Law.
Ideal Gas Law Formula :
General Gas Equation: PV = nRT
Pressure(P) = nRT / V
Volume(V) = nRT / P
Temperature(T) = PV / nR
Moles of Gas(n) = PV / RT where, P = pressure,
V = volume, n = moles of gas,
T = temperature,
R = 8.314 J K-1 mol-1, ideal gas constant.
Ideal Gas Law Example:
Case 1: Find the volume from the 0.250 moles gas at 200kpa and 300K temperature. P = 200 kPa, n = 0.250 mol, T = 300K, R = 8.314 J K-1 mol-1 Step 1: Substitute the values in the below volume equation: Volume(V) = nRT / P
= (0.250 x 8.314 x 300) / 200 = 623.55 / 200 Volume(V) = 3.12 L
This example will guide you to calculate the volume manually.
Case 2: Find the temperature from the 250ml cylinder contaning 0.50 moles gas at 153kpa. V = 250ml -> 250 / 1000 = 0.250 L, n = 0.50 mol, P = 153 kPa, R = 8.314 J K-1 mol-1 Step 1: Substitute the values in the below temperature equation: Temperature(T) = PV / nR = (153 x 0.250) / (0.50 x 8.314) = 38.25 / 4.16 Temperature(T) = 9.2 K
This example will guide you to calculate the temperature manually.
Gay-Lussac's Law: Gay-Lussac's Law states that the pressure of a fixed amount of gas at fixed volume is directly proportional to its temperature in kelvins.
Gay-Lussac's Law Formula :
Gas Equation: Pi/Ti = Pf / Tf
Initial Pressure(Pi) = PfTi / Tf
Initial Temperature(Ti) = PiTf / Pf
Final Pressure(Pf) = PiTf / Ti
Final Temperature(Tf) = PfTi / Pi where, Pi = Initial Pressure,
Ti = Initial Temperature,
Pf = Final Pressure. Tf = Final Temperature,
Gay-Lussac's Law Example:
Case 1: A cylinder contain a gas which has a pressure of 125kPa at a temperature of 200 K. Find the temperature of the gas which has a pressure of 100 kPa. Pi = 125 kPa, Ti = 200 K, Pf = 100
Ideal Gas Law Formula :
General Gas Equation: PV = nRT
Pressure(P) = nRT / V
Volume(V) = nRT / P
Temperature(T) = PV / nR
Moles of Gas(n) = PV / RT where, P = pressure,
V = volume, n = moles of gas,
T = temperature,
R = 8.314 J K-1 mol-1, ideal gas constant.
Ideal Gas Law Example:
Case 1: Find the volume from the 0.250 moles gas at 200kpa and 300K temperature. P = 200 kPa, n = 0.250 mol, T = 300K, R = 8.314 J K-1 mol-1 Step 1: Substitute the values in the below volume equation: Volume(V) = nRT / P
= (0.250 x 8.314 x 300) / 200 = 623.55 / 200 Volume(V) = 3.12 L
This example will guide you to calculate the volume manually.
Case 2: Find the temperature from the 250ml cylinder contaning 0.50 moles gas at 153kpa. V = 250ml -> 250 / 1000 = 0.250 L, n = 0.50 mol, P = 153 kPa, R = 8.314 J K-1 mol-1 Step 1: Substitute the values in the below temperature equation: Temperature(T) = PV / nR = (153 x 0.250) / (0.50 x 8.314) = 38.25 / 4.16 Temperature(T) = 9.2 K
This example will guide you to calculate the temperature manually.
Gay-Lussac's Law: Gay-Lussac's Law states that the pressure of a fixed amount of gas at fixed volume is directly proportional to its temperature in kelvins.
Gay-Lussac's Law Formula :
Gas Equation: Pi/Ti = Pf / Tf
Initial Pressure(Pi) = PfTi / Tf
Initial Temperature(Ti) = PiTf / Pf
Final Pressure(Pf) = PiTf / Ti
Final Temperature(Tf) = PfTi / Pi where, Pi = Initial Pressure,
Ti = Initial Temperature,
Pf = Final Pressure. Tf = Final Temperature,
Gay-Lussac's Law Example:
Case 1: A cylinder contain a gas which has a pressure of 125kPa at a temperature of 200 K. Find the temperature of the gas which has a pressure of 100 kPa. Pi = 125 kPa, Ti = 200 K, Pf = 100