Mr.Luk - Gases · 2019-09-21 · Ideal*Gas*Law* • Combining$the$three$ laws$and$Avogadro’s$...
Transcript of Mr.Luk - Gases · 2019-09-21 · Ideal*Gas*Law* • Combining$the$three$ laws$and$Avogadro’s$...
Avogadro’s Law • “At constant pressure and temperature, the volume of gas is proportional to the quantity (moles) of gas.” =constant
• When quantity increases, volume increases
• When quantity decreases, volume decreases
• At constant P and T,
V ∝nnV
V1n1=V2n2
Avogadro’s Law • You have been using an application of Avogadro’s Law in stoichiometric calculations: Molar volume at STP • At STP (1 atm, 0oC)
1 mol of gas = 22.4* L of gas *Note on STP: The de.inition of STP has changed from 1 atm, 0oC to 100kPa, 0oC (1982) Using this new de.inition
1 mol of gas = 22.7 L of gas
Ideal Gas Law • Combining the three laws and Avogadro’s Law, we get the Ideal Gas Law:
• This is the ideal gas law equation. It relates the four quantities that we mentioned at the beginning of this chapter!
V = k11P
V = k2TV = k3n
V = k1k2k3nTP
V = R nTP
PV = nRT
Ideal Gas Law P: pressure (atm, kPa, mmHg, or torr) V: volume (L) n: quantity (mol) T: temperature (K) R: Gas constant = 0.08026,
= 8.314, or = 62.36
Ideal Gas Law • The ideal gas law equation can be used to derive all of the previous laws • Like before, any variable that is constant can be crossed out to obtain the previously learned gas laws
• Moreover, it can be used in mole and stoichiometric calculations (yes!!!!)
P1V1n1T1
=P2V2n2T2
Ideal Gas Law • The ideal gas law has another form that is helpful for determining the molar mass of the gas based on density of the gas • Regular form: 𝑷𝑽=𝒏𝑹𝑻 • Molar mass/Density form: (𝑴𝒐𝒍𝒂𝒓 𝑴𝒂𝒔𝒔)= 𝒅𝑹𝑻/𝑷 • You will derive this in the homework J
Ideal Gas Law • Example:
v A sealed steel container (hint: constant V and n) contains a gas at 500oC and 2.5 atm. If the container is cool to 300oC, what is the new pressure?
v Show using the ideal gas law that 1.00 mol of gas at STP takes up 22.4 L of space (use 1 atm and 0oC)
v How much space in liters does 1.50 mol of gas take up at 3.0 atm and room temperature?
Ideal Gas Law • Example:
v A 100.0 gram sample of acetone (C3H6O) is vapourized at 100.0 kPa and 28oC. What volume of gas does the vapor take in liters?
v Lab Technique Question: 37.4 grams of a volatile liquid (toluene) is vapourized at 2.0 atm and 300K. The vapours are collected and measured to be 5.0 L. What is the molar mass of the liquid?
v What is the molar mass of a gas that has a density of 1.428 g/L at 0oC and 1.00 atm? (STP) What is the formula/identity of that gas?
Stoichiometry • Instead of solving for quantities only at STP, we can now solve for quantities at any pressure and temperature using the ideal gas law • Example:
v For the reaction 2 C(s) + O2(g) à 2 CO(g), what volume of CO(g) is produced when 20.0g of C reacted at 150 kPa and 100oC ?
g of C à mol of C à mol of CO à L of CO
Stoichiometry • Example:
v C2H4(g) + 3 O2(g) à 2 CO2(g) + 2 H2O(g) When 10.0 L of C2H4(g) reacts with excess O2(g) under 10.0 atm and 600K, what volume of CO2 and H2O are produced? (Note: Do you really need the pressure and temperature? Why or why not?) v 2 H2O(l) à 2 H2(g) + O2(g) The decomposition of water produces two gases. How many litres of gas are produced when 180.g of water is decomposed at 1.00 atm and 298K?
Quiz 2
• Describe how Boyle’s, Charles’, or Gay-‐Lussac’s law works using ideal gas law • Use the above law quantitatively • Determine molar mass of gas using ideal gas law • Stoichiometry calculation
Revisi=ng Vapour Pressure • Why do liquid molecules escape into the gas phase even though the temperature is not equal to the boiling point?
• At any temperature, the energy of individual molecules following a distribution (known as Boltzmann’s Distribution). Not all molecules have the same energy.
• At higher temperatures, the average energy is higher, but that does not mean ALL molecules have higher energy than the ones at a lower temperature
Revisi=ng Vapour Pressure • At higher temperature, the max KE is greater (wider curve), and the average KE is
greater (peak to the right) • Yet the # of molecules (area under the curve) remains the same as lower temperature. • This mean the height of the curve must be lower.
Revisi=ng Vapour Pressure • Notice that at a higher temperature, the number of molecules with enough energy to vapourize (escape as gas) is greater than the number of molecules able to do so at a lower temperature.
Revisi=ng Vapour Pressure • This conkirms what we noted earlier in the Bonding unit: v At a higher temperature, more molecules have enough energy to escape into the gas state
v More molecules in the gas state means greater pressure (PV = nRT)
v Higher temperatures result in higher vapour pressures for all substances
Vapour Pressure and Boiling Point • Let us look at some data for H2O
• What do we notice about the vapour pressure and temperature in the last row?
Temperature (oC) Vapour Pressure (kPa)
0 0.61 20 2.34 40 7.37 60 19.9 80 47.3 100 101.3
Vapour Pressure and Boiling Point • The boiling point of a substance is the temperature at which
vapour pressure = atmospheric pressure • At higher altitudes, when atmospheric pressure is lower (such as on mountains), the boiling point of substances decrease v On the peak of Mt. Everest, water boils at 70oC… this just means you cannot get liquid water about 70oC since it will boil away quickly. Can’t make a good cup of coffee on Everest L Patm ~ 250 torr.