DO NOW: 1)What are the four variables used to describe a gas

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DO NOW: 1)What are the four variables used to describe a gas DO NOW: 1)What are the four variables used to describe a gas? 2) If the volume of a gas decreased, the pressure of the would (increase/decrease)

Boyle’s and Charles’s Laws Day 2 – Notes Unit: Gas Laws Boyle’s and Charles’s Laws

After today you will be able to… Explain the effect on gas properties using Boyle’s and Charles’s laws Calculate an unknown pressure, temperature, or volume by solving algebraically

Boyle’s Law: Pressure and Volume Robert Boyle was the first person to study the pressure-volume relationship of gases. In 1662 Boyle proposed a law to describe this relationship.

“For a given mass of a gas at constant temperature, the volume of a gas varies inversely with pressure.”

Boyle’s Law: Pressure and Volume We can simplify this relationship by the formula: Where, P1, P2 = pressure in any unit (atm, kPa, mmHg), BUT they must match! V1, V2 = volume in any unit (usually, L or mL), BUT they must match! P1 V1 = P2 V2

Boyle’s Law: Example P1 V1=P2 V2 V2 180L = A gas has a volume of 30.0L at 150 kPa. What is the volume of the gas at 0.252 atm? P1= V1= P2= V2= 1 atm 150 kPa x P1 V1=P2 V2 = 1.5 atm 101.3 kPa 30.0L 0.252 atm ? (1.5 atm) (30.0L) = (0.252 atm) (V2) V2 = 180L

Charles’s Law: Temperature and Volume Jacques Charles studied the effect of temperature on volume of a gas at constant pressure. In 1787 Charles proposed a law to describe his observations.

“For a given mass of a gas, at constant pressure, the volume of the gas is directly proportional to its temperature in Kelvin.”

DEMONSTRATION!

Charles’s Law: Temperature and Volume We can simplify this relationship by the formula: Where, V1, V2 = volume in any unit (L or mL), BUT they must match! T1, T2 = temperature is always in Kelvin! (Recall, just add 273 + °C) V1 V2 T1 T2 =

Charles’s Law: Example A gas has a volume of 4.0L at 27°C. What is its volume at 153°C? V1= T1= V2= T2= V1 V2 T1 T2 4.0L = 27°C +273= 300K ? 153°C+273= 426K (4.0L) (V2) = (300K) (426K) V2 = 5.7L

DEMONSTRATION!

Complete your worksheet!

Questions? Complete WS 2