1 5. The general ideal gas equation Gases often experience transformations in which no quantity remains constant. We can establish the relationship between all of them by combining the three laws. Let’s assume that the gas goes from state 1 to state 2, passing through intermediate state A. It does so in this way : 15 Look at the partially filled bottle of water flying at 10 000 m inside a plane at a temperature of 20°C. What will the bottle look like when it lands at an airport where the temperature is 40 ºC? State 1 Transformation State A Transformation State 2 p1 T1 = TA = constant pA pA = p2 = constant p2 V1 Boyle-Mariotte law VA Charles' law V2 T1 p1 × V1 = pA × VA TA V T V T A A = 2 2 T2 We know that T1 = TA and pA = p2. Making VA the subject: V p V p V V T T p V V p T T A A A A = ⋅ = ⋅ → ⋅ = ⋅ 1 1 2 2 1 1 2 2 1 2 ; General ideal gas equation It relates the p, V and T of a gas in any two states. p V T p V T 1 1 1 2 2 2 × × = 16 A cylinder with a movable piston contains 3 L of nitrogen gas at 900 hPa when the temperature is 25 °C. What will the temperature be if the manometer reads 2 atm and the volume is 1.5 L? 17 A cylinder with a movable piston contains 3 L of nitrogen gas at 900 hPa when the temperature is 25 °C. What pressure will the manometer show if the thermometer reads -20 °C and the volume is 1.5 L? A C T I V I T I E S SOLVED PROBLEM 5 On a plane at 20 °C, we have a bottle partially filled with water. In the free space of the bottle, 200 mL is occupied by a gas at a pressure of 750 hPa. On landing, the temperature is 40 °C and the pressure is 1 atm. What volume does the gas in the bottle occupy? 1. Write the two states of the gas with their quantities. p1 = 750 hPa V1 = 200 mL T1 = 20 °C p2 = 1 atm V2 = ? T2 = 40 °C 2. Check that no quantity remains constant. Then use the ideal gas equation. p T p T 1 1 1 2 2 2 ⋅ = ⋅ V V 3. M ake V2 the subject. p T p T p T p T 1 1 1 2 2 2 1 1 2 2 2 1 ⋅ = ⋅ → ⋅ ⋅ = ⋅ ⋅ V V V V ® V p V p T T 2 1 1 2 1 2 = ⋅ ⋅ ⋅ 4. B efore substituting the data, write all the amounts of each quantity in the same units. • Change the units of pressure: 1 hPa 1 hPa atm atm ⋅ = 1013 25 1013 . • The temperature must be in kelvin: T1 (K) = T1 (°C) + 273 = 20 °C + 273 = 293 K T2 (K) = T2 (°C) + 273 = 40 °C + 273 = 313 K V V 2 1 1 2 1 2 = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ = p p T T 750 hPa 200 mL 1013 hPa 293 K 313 K 158 mL The increase in temperature causes the volume to decrease less than in solved problem 2. CHALLENGE 19
RkJQdWJsaXNoZXIy