Combined Gas Law and Avogadros Hypothesis Combined Gas

Combined Gas Law and Avogadro’s Hypothesis

Combined Gas Law • Recall…. • P x V = constant And V/T = constant And P/T = constant It follows then that……

Combined Gas Law • The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. P 1 V 1 P 2 V 2 = T 1 T 2

Combined Gas Law P 1 V 1 T 1 = P 2 V 2 T 2 Rearrange the combined gas law to solve for V 2 P 1 V 1 T 2 V 2 = P 1 V 1 T 2 P 2 T 1 = P 2 V 2 T 1

Solution C 1 Solve the combined gas law for T 2. (Hint: cross-multiply first. ) P 1 V 1 T 1 = P 2 V 2 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 P 1 V 1

Combined Gas Law Problems A sample of helium gas has a volume of 0. 180 L, a pressure of 0. 800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90. 0 m. L and a pressure of 3. 20 atm? Set up Data Table P 1 = 0. 800 atm V 1 = 0. 180 L P 2 = 3. 20 atm V 2= 0. 090 L T 1 = 302 K T 2 = ? ?

Calculation P 1 = 0. 800 atm P 2 = 3. 20 atm P 1 V 1 T 1 = V 1 = 180 m. L V 2= 90 m. L P 2 V 2 T 1 = 302 K T 2 = ? ? P 1 V 1 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = 3. 20 atm x 90. 0 m. L x 302 K 0. 800 atm x 180. 0 m. L T 2 = 604 K - 273 = 331 °C = 604 K

Learning Check C 2 A gas has a volume of 675 m. L at 35°C and 0. 850 atm pressure. What is the temperature in °C when the gas has a volume of 0. 315 L and a pressure of 802 mm Hg?

Solution G 9 T 1 = 308 K T 2 = ? V 1 = 675 m. L V 2 = 0. 315 L = 315 m. L P 1 = 0. 850 atm P 2 = 802 mm Hg = 646 mm Hg T 2 = 308 K x 802 mm Hg 646 mm Hg P inc, T inc = 178 K - 273 = - 95°C x 315 m. L 675 m. L V dec, T dec

Learning Check C 3 True (1) or False(2) 1. ___The P exerted by a gas at constant V is not affected by the T of the gas. 2. ___ At constant P, the V of a gas is directly proportional to the absolute T 3. ___ At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

Solution C 3 True (1) or False(2) 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

Avogadro’s Law When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V a number of moles (n) V = constant x n V 1 n 1 initial = V 2 n 2 final

Avogadro’s Law (continued) • Equal volumes of gases at same temperature and pressure have equal numbers of molecules. • Gas molecules may break up when they react.

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4 NH 3 + 5 O 2 4 mole NH 3 4 NO + 6 H 2 O 4 mole NO At constant T and P 1 volume NH 3 1 volume NO

New Combined Gas Law • Boyle’s law V a 1/P • Charles’s law V a T • Avogadro’s law V a n Therefore it follows that…… V a n. T P

The General Gas Equation P 2 V 2 P 1 V 1 = n 1 T 1 n 2 T 2 If we hold the number of moles and volume constant: P 1 T 1 = P 2 T 2

STP The volumes of gases can be compared when they have the same temperature and pressure (STP). Standard temperature Standard pressure k. Pa 0°C or 273 K 1 atm (760 mm Hg) or 101. 3

Learning Check C 4 A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2. 0 atm and – 25°C? P 1 = P 2 = V 2 = 15 L x V 1 V 2 = = ? atm T 1 T 2 x = = K K K

Solution C 4 P 1 = 1. 0 atm P 2 = 2. 0 atm V 1 = 15 L V 2 = ? ? V 2 = 15 L x 1. 0 atm 2. 0 atm T 1 = 273 K T 2 = 248 K x 248 K = 6. 8 L 273 K

Molar Volume At STP 4. 0 g He 1 mole (STP) 16. 0 g CH 4 1 mole (STP) V = 22. 4 L 44. 0 g CO 2 1 mole (STP) V = 22. 4 L

Molar Volume Factor 1 mole of a gas at STP = 22. 4 L 1 mole and 1 mole 22. 4 L

Learning Check C 5 A. What is the volume at STP of 4. 00 g of CH 4? 1) 5. 60 L 2) 11. 2 L 3) 44. 8 L B. How many grams of He are present in 8. 0 L of gas at STP? 1) 25. 6 g 2) 0. 357 g 3) 1. 43 g

Solution C 5 A. What is the volume at STP of 4. 00 g of CH 4? 4. 00 g CH 4 x 1 mole CH 4 x 22. 4 L (STP) = 5. 60 L 16. 0 g CH 4 1 mole CH 4 B. How many grams of He are present in 8. 0 L of gas at STP? 8. 00 L x 1 mole He x 4. 00 g He = 1. 43 g He 22. 4 He 1 mole He