The combined gas law

1. The combined gas law Putting It All Together

2. Objectives • When you complete this presentation, you will be able to • state the combined gas law • relate the combined gas law to Boyle’s law, Charles’s law, and Gay-Lussac’s law • use the combined gas law to compute pressure, volume, and temperature values for a gas system.

3. Introduction Temperature is constant • Boyle’sLaw: P1×V1 = P2×V2 • Charles’sLaw: V1/T1 =V2/T2 • Gay-Lussac’s Law: P1/T1 = P2/T2 • Theseonlywork if weholdsomething constant. Pressure is constant Volume is constant

4. Introduction • What happens if we combine the three gas laws and don’t keep anything constant? • The combined gas law: P1 × V1 P2 × V2 = T1 T2

5. Application • If we want to solve for one variable, we need to know the values of the other 5 variables. P2×V2 P1×V1 P1 = × × × × × × P2 = V1 = V2 = T2 P2 × V2 T2 P1 × V1 T2 T1 T1 P1 × V1 P2 × V2 T1 V1 P2 × V2 P2 T2 T1 T2 P1× V1 P1 T1 V2 T1 = T2 =

6. Application • If we want to solve for one variable, we need to know the values of the other 5 variables. P1 = P2 = V1 = V2 = P1 × V1 × T2 T2 × P1 × V1 P2 × V2 × T1 P1 × V1 × T2 P2 × V2 × T1 T1 × P2 × V2 T1 × P2 T1 × V2 P1 × V1 T2 × P1 T2 × V1 P2 × V2 T1 = T2 =

7. Example 1 – Finding P2 A 542 mL sample of H2 gas has a temperature of 300 K and a pressure of 102 kPa. What is the pressure of the gas when the volume is 750 mL and the temperature is 310 K? P1 = 102 kPa P2 = ? kPa V1 = 542 mL V2 = 750 mL T1 = 300 K T2 = 310 K = 76.2 kPa = 76.16906667 kPa P2 = = (102 kPa)(542 mL)(310 K) P1 × V1 × T2 T1 × V2 (300 K)(750 mL) Is this reasonable? T increases a little (by a factor of about 0.03) which means P increases by a factor of about 0.03. V increases a lot (by a factor of about 1.5) which means P decreases by a factor of about 1.5. This means that P2 should be about P1/1.5 = 102/1.5 ≈ 200/3 ≈ 67 (which is within 15% of 76.2). The answer is reasonable.

8. Example 2 – Finding V2 A 4.00 L sample of Ne gas has a temperature of 623 K and a pressure of 0.250 atm. What is the volume of the gas when the pressure is 1.00 atm and the temperature is 273 K? P1 = 0.250 atm P2 = 1.00 atm V1 = 4.00 L V2 = ? L T1 = 623 K T2 = 273 K = 0.438 L = 0.4382022472 L V2 = = (0.250 atm)(4.00 L)(273 K) P1 × V1 × T2 T1 × P2 (623 K)(1.00 atm) Is this reasonable? T decreases a lot (to about 40% of the original value) which means V decreases to about 40%. P increases a lot (by about 400%) which means V decreases to about 25% of the original value. 40% times 25% is about 10%; 0438 L is about 10% of 4.00 L. It is reasonable.

9. Example 3 – Finding T2 A balloon is filled with 525 L of He gas at a temperature of 310 K and a pressure of 0.995 atm. What is the temperature of the gas when the pressure is 0.250 atm and the volume is 1620 L? P1 = 0.995 atm P2 = 0.250 atm V1 = 525 L V2 = 1620 L T1 = 310 K T2 = ? K = 240 K = 240.34458 K T2= = (310 K)(0.250 atm)(1620 L) T1× P2× V2 P1× V1 (0.995 atm)(525 L) Is this reasonable? P decreases a lot (to about 25% of the original value) which means T decreases to about 25% of the original value. V increases a lot (to about 300% the original value) which means T increases about 300% the original value. 25% times 300% is about 75%; 240 K is about 75% of 310 K. It is reasonable.

10. Summary • The combined gas law is a combination of Boyle’s law, Charles’s law, and Gay-Lussac’s law. • The equation: P1 × V1 P2 × V2 = T1 T2