Book 1 Section 5.1 The gas laws

Book 1 Section 5.1 The gas laws
The inflation of snack packages Gas pressure Check-point 1 Boyle’s law Check-point 2 The pressure law Check-point 3 Charles’ law Check-point 4 The general gas law and ideal gases Check-point 5 1 2 3 4 5 Book 1 Section 5.1 The gas laws

The inflation of snack packages
A pack of snacks was found to be inflated at Huanglong (altitude of over 3000 m). What is the physics behind this? Pressure. Book 1 Section 5.1 The gas laws

1 Gas pressure a What is pressure? When juice is removed, the box collapses pressure inside the box pressure from the gas outside << Video 5.1 Collapsing can Book 1 Section 5.1 The gas laws

a What is pressure? Pressure = normal force on a surface area of the surface F A = symbol : p unit : pascal (Pa) 1 Pa = 1 N m-2 Bed of nails Example 1 Book 1 Section 5.1 The gas laws

1 Gas pressure b Atmospheric pressure Atmospheric pressure: pressure from the atmosphere around us (~100 kPa). acts on all surfaces exposed to the atmosphere, including our bodies. balanced by an equal pressure within our body. Book 1 Section 5.1 The gas laws

1 Gas pressure c Measuring gas pressure Gas pressure is measured using a Bourdon gauge. Curved metal tube uncoils when pressure is applied  moves the pointer around a dial Book 1 Section 5.1 The gas laws

Check-point 1 – Q1 In a wooden block game, there are 48 wooden blocks. Dimensions of each block: 1.5  7.5  2.5 cm Mass of each block: 24 g A block is picked from the bottom level and put on the top. Average pressure exerted on the floor by the blocks = ? Book 1 Section 5.1 The gas laws

Check-point 1 – Q1 Weight of the whole set = mg = 24  48  10–3  10 = N Contact area = 2.5  7.5  2 = 37.5 cm2 F A = 11.52 37.5  10–4 Pressure = = 3072 Pa Book 1 Section 5.1 The gas laws

Check-point 1 – Q2 Bourdon gauge The apparatus shown is a ________________. It measures __________. The pointer points to 1.0. This shows that the ___________ is about __________ kPa. pressure pressure 100 Book 1 Section 5.1 The gas laws

2 Boyle’s law Quantities like pressure, volume, temp and mass are usually used to describe gases. Boyle’s law: Relationship between pressure and volume of gas of fixed mass and constant temp. Simulation 5.1 Boyle’s law Boyle’s law Expt 5a Book 1 Section 5.1 The gas laws

2 Boyle’s law p -1/V graph: gives a straight line passing through the origin. p -V graph: gives a curve. Book 1 Section 5.1 The gas laws

2 Boyle’s law Boyle’s law: For a gas with a fixed mass and temperature, its pressure is inversely proportional to its volume, i.e. p  (constant T ) 1 V pV = constant or p1V1 = p2V2 (constant T )  Gas guns Example 2 Book 1 Section 5.1 The gas laws

2 Boyle’s law Video 5.3 Growing marshmallow Book 1 Section 5.1 The gas laws

2 Boyle’s law The physics behind : At Huanglong (altitude > 3000 m), atmosphere pressure is lower than that at sea level the gas pressure inside the package tends to attain the atmosphere pressure volume of the trapped air inside the package  Book 1 Section 5.1 The gas laws

Check-point 2 – Q1 A syringe contains 50 cm3 of air at atmospheric pressure (100 kPa). Assume Boyle’s law holds. What is the condition on the temperature for Boyle’s law to hold? Temperature remains unchanged. Book 1 Section 5.1 The gas laws

Check-point 2 – Q1 (b) If the volume is reduced to 45 cm3, what will be the new pressure? By Boyle’s law, p1V1 V2 p2 = = 100  1000  50 45 = 111 kPa Book 1 Section 5.1 The gas laws

Check-point 2 – Q2 In an experiment using a gas of constant mass and temperature to study Boyle’s law, a p-V graph is drawn: Book 1 Section 5.1 The gas laws

Check-point 2 – Q2 (a) What is the volume of the gas if the pressure is 120 kPa? 120 0.01 When p = 120 kPa, V = m3 Book 1 Section 5.1 The gas laws

Check-point 2 – Q2 (b) Verify Boyle’s law using the information given in the graph. Let p1 = 150 kPa, p2 = 120 kPa At p1, V1 = m3 At p2, V2 = 0.01 m3 p1V1 = 150  1000  = 1200 N m p2V2 = 120  1000  0.01 = 1200 N m = p1V1  pV is constant Book 1 Section 5.1 The gas laws

3 The pressure law a Pressure-temperature relationship of a gas Simulation 5.2 Pressure law Pressure-temperature relationship of a gas Expt 5b Book 1 Section 5.1 The gas laws

a Pressure-temperature relationship of a gas
Expt 5b shows that there is a linear relationship between pressure and temperature. If the graph is extrapolated to the temp axis, cuts the axis at –273 C. –273  Expt with different gases give the same intercept. Book 1 Section 5.1 The gas laws

a Pressure-temperature relationship of a gas
If we shift the pressure axis to the left to the intersecting point, a graph passing through the origin is obtained.  p  (TC + 273) p : pressure of the gas TC : Celsius temperature of the gas Book 1 Section 5.1 The gas laws

3 The pressure law b The Kelvin temperature scale A new temp scale: the Kelvin temperature scale TK (Kelvin temp) = TC (Celsius temp) + 273 Unit of TK: kelvin symbol: K Zero kelvin = absolute zero  Kelvin temp scale is also called the absolute temperature scale Note: ∆T of 1 K = ∆T of 1 C. Book 1 Section 5.1 The gas laws

3 The pressure law c The pressure law For a gas with a fixed mass and volume, its pressure is directly proportional to its Kelvin temperature, i.e. p T (constant V ) (constant V ) = constant or p T p1 T1 = p2 T2  Pressure inside light bulbs Example 3 Book 1 Section 5.1 The gas laws

Check-point 3 – Q1 Complete the following table: C K Body temp 37 ______ Melting point of mercury 234 310 ̶ 39 Book 1 Section 5.1 The gas laws

Check-point 3 – Q2 A sealed container holds a gas at 300 K. For its pressure to triple in value, to which temperature should it be heated? Assume the volume of the gas is fixed. A 100 K C 900 K B 100 C D 900 C Book 1 Section 5.1 The gas laws

Check-point 3 – Q3-4 A student uses a gas with a fixed mass and volume to study the pressure law. The graph is as follows: Book 1 Section 5.1 The gas laws

Check-point 3 – Q3 What is the pressure if the temp is 290 K? A 60 kPa B 72 kPa C 84 kPa D 110 kPa Book 1 Section 5.1 The gas laws

Check-point 3 – Q4 What is the pressure if the temp is 600 K? A kPa B 110 kPa C 149 kPa D 157 kPa Book 1 Section 5.1 The gas laws

4 Charles’ law Simulation 5.3 Charles’ law Volume-temperature relationship of a gas Expt 5c Book 1 Section 5.1 The gas laws

4 Charles’ law The V -T graph is a straight line passing through the origin: Book 1 Section 5.1 The gas laws

4 Charles’ law Charles’ law : For a gas with a fixed mass and pressure, its volume is directly proportional to its Kelvin temperature, i.e. V T (constant p ) (constant p ) = constant or V T V1 T1 = V2 T2  Book 1 Section 5.1 The gas laws

4 Charles’ law Video 5.6 Dented table tennis ball Volume of a balloon Example 4 Book 1 Section 5.1 The gas laws

Check-point 4 – Q1 A student uses the Celsius temp instead of the Kelvin temp in doing Expt 5c and plots a graph: absolute zero Relationship between V and T of the gas = ? V T. (b) Mark the position of absolute zero on the temp axis. Book 1 Section 5.1 The gas laws

5 The general gas law and ideal gases a The general gas law p  (constant T ) 1 V Boyle’s law: Pressure law: p T (constant V ) Charles’ law: V T (constant p ) = constant pV T pV T or  p1V1 T1 = p2V2 T2  Book 1 Section 5.1 The gas laws

a The general gas law V  N (N : number of molecules) 1 mole (mol) = 6.02  1023 molecules Avogadro number (NA) N = nNA  N  n  pV  nT General gas law: pV = nRT R: universal gas constant = 8.31 J mol-1 K-1 Book 1 Section 5.1 The gas laws

5 The general gas law and ideal gases b Ideal gases Ideal gas: gas obeys the general gas law Real gas: does not obey it precisely, especially at high p or low T behaves like an ideal gas at low p and high T Hot air balloon Example 5 Book 1 Section 5.1 The gas laws

Check-point 5 – Q1 A syringe contains 80 cm3 of air at 27 C and 100 kPa. The piston is quickly pushed in to reduce the volume to 70 cm3. Temp  to 33 C. What is the new pressure? p1V1 T1 = p2V2 T2 By = p1V1 T1  T2 V2 p2 = 100  103  80  10-6  70  10-6 =117 kPa Book 1 Section 5.1 The gas laws

Check-point 5 – Q2 Estimate the number of moles of air inside a bedroom at 25 C and atmospheric pressure. Dimensions of bedroom: 2.5 m  2 m  2.2 m. (Take R = 8.31 J mol-1 K-1) By the general gas law, n = pV RT = 105  2.5  2  2.2 8.31  ( ) = 444 mol Book 1 Section 5.1 The gas laws

The End Book 1 Section 5.1 The gas laws

Book 1 Section 5.1 The gas laws

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