Review%20Session-1

# Review%20Session-1 - REVIEW SESSION‐ 1 ...

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Unformatted text preview: REVIEW SESSION‐ 1  GAS LAWS  Conversion of Pressure:  Arrange the pressures in ascending order (1) 30 cm of Hg (2) 30 lb/in2  (3) 0.76 torr  Express the following gas pressures in ‘atm’  1) 30 cm of Hg  2) 30 lb/in2  3) 0.76 torr  Conversion of Pressure:  Arrange the pressures in ascending order (1) 30 cm of Hg (2) 30 lb/in2  (3) 0.76 torr  Express the following gas pressures in ‘atm’  1)  30 cm of Hg: 300 mm of Hg (300 mm) ( 1atm/ 760 mm)                                                 = 0.395 atm  2) 30 lb/in2: (30 lb/in2) (1atm/ 14.7lb/in2)                                                                         = 2.05 atm  3) 0.76 torr : (0.76 torr) (1 atm/ 760 torr) = 0.001 atm  The ascending order: 0.76 torr < 30cm < 30 lb/in2  ‐ The volume of the lungs is measured by the volume of air inhaled  or exhaled.  If the  volume of the lungs is 2.400 L during exhalaQon  and the pressure is 101.70 kPa, and the pressure during inhalaQon is  100.01 kPa, what is the volume of the lungs during inhalaQon?   ‐ The volume of the lungs is measured by the volume of air inhaled  or exhaled.  If the  volume of the lungs is 2.400 L during exhalaQon  and the pressure is 101.70 kPa, and the pressure during inhalaQon is  100.01 kPa, what is the volume of the lungs during inhalaQon?   P1 = 101.7 kPa               V1 = 2.4 L               P2 = 100.01 kPa  V2 = ?  P1V1 = P2V2  V2 = P1V1/P2  V2 = (101.7 kPa) (2.4L) / (100.01 kPa)  V2 = 2.441L  ‐ Does the volume of a ﬁxed quanQty of gas decreases to half its original   value when the temperature is lowered from 100oC to 50oC ?  ‐ Does the volume of a ﬁxed quanQty of gas decreases to half its original   value when the temperature is lowered from 100oC to 50oC ?  ‐ NO  ‐ Does the volume of a ﬁxed quanQty of gas decreases to half its original   value when the temperature is lowered from 100oC to 50oC ?  ‐ NO  V1 / T1 =  V2 / T2  V1 / V2 = T1 / T2  V1 / V2 = (100+273) / (50+273)  V1 / V2 = 1.154  V1 = V2 (1.154)  V2 = (0.87) V1   ‐ Tennis balls are ﬁlled with N2 gas to higher pressures to increase their   bounce. If a parQcular tennis ball has a volume of 144 cm3 and contains   0.33g N2 gas, what is the pressure inside the tennis ball at 24oC   ‐ Tennis balls are ﬁlled with N2 gas to higher pressures to increase their   bounce. If a parQcular tennis ball has a volume of 144 cm3 and contains   0.33g N2 gas, what is the pressure inside the tennis ball at 24oC   V=  144 cm3 = 144 mL = 0.144L  T = 24oC= 273.1 +24 = 297.1 K  n= Wt/ MWt = 0.33g/ 28 (g /mol) = 0.0118 mol  R = 0.08206 L‐atm/Mol. K  P = nRT/V  P= (0.0118 mol) (0.08206 L‐atm/ Mol. K) 297.1 K / 0.144L  P =2 atm    ‐ 6.2 liters of an ideal gas are contained at 3.0 atm and 37 °C. How  many moles of this gas are present?  ‐ 6.2 liters of an ideal gas are contained at 3.0 atm and 37 °C. How  many moles of this gas are present?  Given,  P = 3 atm  V= 6.2 L  T = 273.1+37 = 310.1K  R= 0.08206 L‐atm/mol.K  ‐ 6.2 liters of an ideal gas are contained at 3.0 atm and 37 °C. How  many moles of this gas are present?  Given,  Asking for “n”  P = 3 atm  PV= nRT  V= 6.2 L  n = PV/RT  T = 273.1+37 = 310.1K     = (3 atm) (6.2 L)        (0.08206 L‐atm/mol. K) 310.1 K  R= 0.08206 L‐atm/mol.K  n = 0.731 mol  Solid potassium chlorate (KClO3) decomposes to produce solid  potassium chloride (KCl) and oxygen gas (O2) according to the balanced  chemical equaQon: 2KClO3 (s)  2KCl (s) + 3O2 (g)  What volume of gas, measure at 23.4oC and 75.4 kPa, will be produced  when 32.5 g of potassium chlorate is decomposed?  Solid potassium chlorate (KClO3) decomposes to produce solid  potassium chloride (KCl) and oxygen gas (O2) according to the balanced  chemical equaQon: 2KClO3 (s)  2KCl (s) + 3O2 (g)  What volume of gas, measure at 23.4oC and 75.4 kPa, will be produced  when 32.5 g of potassium chlorate is decomposed?  Given,  P = 75.4kPa   (75.4kPa) (1 atm)/ (101.325kPa) = 0.745 atm  T = 23.4oC = 273.1+23.4 = 296.5 K  n = wt/MWt = (32.5g)/ (122.5g/mol) = 0.265 mol  R= 0.08206 L.atm/mol.K  Solid potassium chlorate (KClO3) decomposes to produce solid  potassium chloride (KCl) and oxygen gas (O2) according to the balanced  chemical equaQon: 2KClO3 (s)  2KCl (s) + 3O2 (g)  What volume of gas, measure at 23.4oC and 75.4 kPa, will be produced  when 32.5 g of potassium chlorate is decomposed?  Given,  P = 75.4kPa   (75.4kPa) (1 atm)/ (101.325kPa) = 0.745 atm  T = 23.4oC = 273.1+23.4 = 296.5 K  32.5g of KClO3 decomposed  ? g of oxygen produced  2 moles of KClO3 produce 3 moles of Oxygen  2(122.55g) of KClO3 produce 3(32g) of oxygen  32.5g of KClO3 produce (3X32) (32.5)/ (2X 122.55) = 12.73g of Oxygen  n = 12.73/32 = 0.398 mol  Solid potassium chlorate (KClO3) decomposes to produce solid  potassium chloride (KCl) and oxygen gas (O2) according to the balanced  chemical equaQon: 2KClO3 (s)  2KCl (s) + 3O2 (g)  What volume of gas, measure at 23.4oC and 75.4 kPa, will be produced  when 32.5 g of potassium chlorate is decomposed?  Given,  P = 75.4kPa   (75.4kPa) (1 atm)/ (101.325kPa) = 0.745 atm  T = 23.4oC = 273.1+23.4 = 296.5 K  32.5g of KClO3 decomposed  ? g of oxygen produced  2 moles of KClO3 produce 3 moles of Oxygen  2(122.55g) of KClO3 produce 3(32g) of oxygen  32.5g of KClO3 produce (3X32) (32.5)/ (2X 122.55) = 12.73g of Oxygen  n = 12.73/32 = 0.398 mol  V= (0.398) (0.08206) (296.5) / 0.745 = 13 L  Nitrogen and hydrogen gases react to form ammonia gas.           N2(g) + 3 H2(g) → 2 NH3(g)  At a certain temperature and pressure, 2 L of N2 reacts with 7 L of H2.  What volume of NH3 is produced? What gases remain at the end of the   reacQon?  Nitrogen and hydrogen gases react to form ammonia gas.           N2(g) + 3 H2(g) → 2 NH3(g)  At a certain temperature and pressure, 2 L of N2 reacts with 7 L of H2.  What volume of NH3 is produced? What gases remain at the end of the   reacQon?   1 mole of  N2 needs 3 moles of  H2to react completely & produce  2 moles of NH3.   N2 and H2 react with each other in 1:3 raQo  Nitrogen and hydrogen gases react to form ammonia gas.           N2(g) + 3 H2(g) → 2 NH3(g)  At a certain temperature and pressure, 2 L of N2 reacts with 7 L of H2.  What volume of NH3 is produced? What gases remain at the end of the   reacQon?   1 mole of  N2 needs 3 moles of  H2to react completely & produce  2 moles of NH3.   N2 and H2 react with each other in 1:3 raQo   If you have 2 L of N2, you need at least 6 L of H2.  1(2L) of N2 + 3 (2L) H2  2(2L) NH3  So, 4L of NH3 will be produced and 1L of H2 (unreacted) will  remain in the container along with 4L of NH3.  What is the total pressure exerted by a mixture of 2.0g of H2 and 8.0g of  N2 at 273K in a 10.0 L vessel?  What is the total pressure exerted by a mixture of 2.0g of H2 and 8.0g of  N2 at 273K in a 10.0 L vessel?  Given,  T = 273 K  V = 10 L  R = 0.08206 L.atm/mol.K  n = ?  and P =?  What is the total pressure exerted by a mixture of 2.0g of H2 and 8.0g of  N2 at 273K in a 10.0 L vessel?  Given,  T = 273 K  V = 10 L  R = 0.08206 L.atm/mol.K  n = ?  and P =?  n = Number of moles= nH2 +nN2  nH2 = wt/Mwt = (2.0g)/(2.0g/mol) = 1 mol  nN2 = wt/Mwt = (8.0g)/(28.0g/mol)= 0.286 mol  n = 1+ 0.286 =1.286 mol  What is the total pressure exerted by a mixture of 2.0g of H2 and 8.0g of  N2 at 273K in a 10.0 L vessel?  Given,  T = 273 K  V = 10 L  R = 0.08206 L.atm/mol.K  n = Number of moles= nH2 +nN2  nH2 = wt/Mwt = (2.0g)/(2.0g/mol) = 1 mol  nN2 = wt/Mwt = (8.0g)/(28.0g/mol)= 0.286 mol  n = 1+ 0.286 =1.286 mol  n = ?  and P =?  P =nRT/V     = 1.286 X 0.08206 X 273 /10    P = 2.88 atm  In the ﬁrst step of making nitric acid, ammonia and oxygen reacts  to  form nitric oxide as shown in the equaQon:       4NH3(g) + 5O2(g)  4NO(g) + 6 H2O(g)  How many liters of NH3(g) at 850oC and 5atm are required to react with   1.0 mol of O2(g) in this reacQon?  In the ﬁrst step of making nitric acid, ammonia and oxygen reacts  to  form nitric oxide as shown in the equaQon:       4NH3(g) + 5O2(g)  4NO(g) + 6 H2O(g)  How many liters of NH3(g) at 850oC and 5atm are required to react with   1.0 mol of O2(g) in this reacQon?  First, we have to calculate the volume occupied by Oxygen under the  given condiQons.  P = 5atm  T= 850+273 =1123K  n= 1, R= 0.08206 L.atm/mol.K  V = nRT/P = 18.43L  In the ﬁrst step of making nitric acid, ammonia and oxygen reacts  to  form nitric oxide as shown in the equaQon:       4NH3(g) + 5O2(g)  4NO(g) + 6 H2O(g)  How many liters of NH3(g) at 850oC and 5atm are required to react with   1.0 mol of O2(g) in this reacQon?  First, we have to calculate the volume occupied by Oxygen under the  given condiQons.                   Now, look at the equaQon  P = 5atm              5 moles of Oxygen required to react                     with 4 moles of Ammonia.        T= 850+273 =1123K                           means, 5 (22.4 L) of Oxygen requires   n= 1, R= 0.08206 L.atm/mol.K      4(22.4L) of Nitrogen.                       means, 112L of Oxygen requires    V = nRT/P = 18.43L        89.6 L of ammonia.  In the ﬁrst step of making nitric acid, ammonia and oxygen reacts  to  form nitric oxide as shown in the equaQon:       4NH3(g) + 5O2(g)  4NO(g) + 6 H2O(g)  How many liters of NH3(g) at 850oC and 5atm are required to react with   1.0 mol of O2(g) in this reacQon?  First, we have to calculate the volume occupied by Oxygen under the  given condiQons.                   Now, look at the equaQon  P = 5atm              5 moles of Oxygen required to react                     with 4 moles of Ammonia.        T= 850+273 =1123K                           means, 5 (22.4 L) of Oxygen requires   n= 1, R= 0.08206 L.atm/mol.K      4(22.4L) of Nitrogen.                       means, 112L of Oxygen requires    V = nRT/P = 18.43L        89.6 L of ammonia.  So how 18.43 liters of O2 requires ‐‐‐‐‐ Liters of ammonia? 14.75L  A container has 1g of H2, 4g of O2 and 7g of N2 gases at 12 atm pressure.  Calculate the parQal pressure exerted by individual gases?  A container has 1g of H2, 4g of O2 and 7g of N2 gases at 12 atm pressure.  Calculate the parQal pressure exerted by individual gases?  First, Calculate the mole fracQons  need number of moles of each gas  Number of moles   nH2 = wt/Mwt = 1/2 = 0.5 mols  nO2 = wt/Mwt = 4/32 = 0.125 mols  nN2 = wt/Mwt = 7/28 = 0.25 mols  n = 0.5+0.125+0.25 = 0.875 mols  A container has 1g of H2, 4g of O2 and 7g of N2 gases at 12 atm pressure.  Calculate the parQal pressure exerted by individual gases?  First, Calculate the mole fracQons  need number of moles of each gas  Number of moles   nH2 = wt/Mwt = 1/2 = 0.5 mols  Mole fracQons   XH2 = 0.5/0.875 = 0.571  nO2 = wt/Mwt = 4/32 = 0.125 mols  XO2 = 0.125/0.875= 0.143  nN2 = wt/Mwt = 7/28 = 0.25 mols  XN2 = 0.25/0.875 = 0.286  n = 0.5+0.125+0.25 = 0.875 mols  X = 0.571+0.143+0.286 =1.0   A container has 1g of H2, 4g of O2 and 7g of N2 gases at 12 atm pressure.  Calculate the parQal pressure exerted by individual gases?  First, Calculate the mole fracQons  need number of moles of each gas  Number of moles   nH2 = wt/Mwt = 1/2 = 0.5 mols  Mole fracQons   XH2 = 0.5/0.875 = 0.571  ParQal Pressures  PH2 = XH2P = 0.571X12 = 6.84atm  nO2 = wt/Mwt = 4/32 = 0.125 mols  XO2 = 0.125/0.875= 0.143  PO2 = XO2P = 0.143X12 = 1.72atm  nN2 = wt/Mwt = 7/28 = 0.25 mols  XN2 = 0.25/0.875 = 0.286  PN2 = XN2P = 0.286X12 = 3.44atm  n = 0.5+0.125+0.25 = 0.875 mols  X = 0.571+0.143+0.286 =1.0   ...
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