Explanation:
Part A:
Total pressure of the mixture = P = 5.40 atm
Volume of the container = V = 10.0 L
Temperature of the mixture = T = 23°C = 296.15 K
Total number of moles of gases = n
PV = nRT (ideal gas equation)
![n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol](https://tex.z-dn.net/?f=n%3D%5Cfrac%7BPV%7D%7BRT%7D%3D%5Cfrac%7B5.40%20atm%5Ctimes%2010.0%20L%7D%7B0.0821%20atm%20L%2Fmol%20K%5Ctimes%20296.15%20K%7D%3D2.22%20mol)
Moles of methane gas = ![n_1=\frac{8.00 g}{16 g/mol}=0.5 mol](https://tex.z-dn.net/?f=n_1%3D%5Cfrac%7B8.00%20g%7D%7B16%20g%2Fmol%7D%3D0.5%20mol)
Moles of ethane gas =![n_2=\frac{18.0 g}{30 g/mol}=0.6 mol](https://tex.z-dn.net/?f=n_2%3D%5Cfrac%7B18.0%20g%7D%7B30%20g%2Fmol%7D%3D0.6%20mol)
Moles of propane gas = ![n_3](https://tex.z-dn.net/?f=n_3)
![n=n_1+n_2+n_3](https://tex.z-dn.net/?f=n%3Dn_1%2Bn_2%2Bn_3)
![2.22=0.5 mol +0.6 mol+ n_3](https://tex.z-dn.net/?f=2.22%3D0.5%20mol%20%2B0.6%20mol%2B%20n_3)
![n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol](https://tex.z-dn.net/?f=n_3%3D%202.22%20mol%20-%200.5%20mol%20-0.6%20mol%3D%201.12%20mol)
Mole fraction of methane =![\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}](https://tex.z-dn.net/?f=%5Cchi_1%3D%5Cfrac%7Bn_1%7D%7Bn_1%2Bn_2%2Bn_3%7D%3D%5Cfrac%7Bn_1%7D%7Bn%7D)
![\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252](https://tex.z-dn.net/?f=%5Cchi_1%3D%5Cfrac%7B0.5%20mol%7D%7B2.22%20mol%7D%3D0.2252)
Similarly, mole fraction of ethane and propane :
![\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703](https://tex.z-dn.net/?f=%5Cchi_2%3D%5Cfrac%7Bn_2%7D%7Bn%7D%3D%5Cfrac%7B0.6%20mol%7D%7B2.22%20mol%7D%3D0.2703)
![\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045](https://tex.z-dn.net/?f=%5Cchi_3%3D%5Cfrac%7Bn_3%7D%7Bn%7D%3D%5Cfrac%7B1.12%20mol%7D%7B2.22%20mol%7D%3D0.5045)
Partial pressure of each gas can be calculated by the help of Dalton's' law:
Partial pressure of methane gas:
![p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm](https://tex.z-dn.net/?f=p_1%3DP%5Ctimes%20%5Cchi_1%3D5.40%20atm%5Ctimes%200.2252%3D1.22%20atm)
Partial pressure of ethane gas:
![p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm](https://tex.z-dn.net/?f=p_2%3DP%5Ctimes%20%5Cchi_2%3D5.40%20atm%5Ctimes%200.2703%3D1.46%20atm)
Partial pressure of propane gas:
![p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm](https://tex.z-dn.net/?f=p_3%3DP%5Ctimes%20%5Cchi_3%3D5.40%20atm%5Ctimes%200.5045%3D2.72%20atm)
Part B:
Suppose in 100 grams mixture of nitrogen and oxygen gas.
Percentage of nitrogen = 37.8 %
Mass of nitrogen in 100 g mixture = 37.8 g
Mass of oxygen gas = 100 g - 37.8 g = 62.2 g
Moles of nitrogen gas = ![n_1=\frac{37.8 g g}{28g/mol}=1.35 mol](https://tex.z-dn.net/?f=n_1%3D%5Cfrac%7B37.8%20g%20g%7D%7B28g%2Fmol%7D%3D1.35%20mol)
Moles of oxygen gas =![n_2=\frac{62.2 g}{32 g/mol}=1.94 mol](https://tex.z-dn.net/?f=n_2%3D%5Cfrac%7B62.2%20g%7D%7B32%20g%2Fmol%7D%3D1.94%20mol)
Mole fraction of nitrogen=![\chi_1=\frac{n_1}{n_1+n_2}](https://tex.z-dn.net/?f=%5Cchi_1%3D%5Cfrac%7Bn_1%7D%7Bn_1%2Bn_2%7D)
![\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103](https://tex.z-dn.net/?f=%5Cchi_1%3D%5Cfrac%7B1.35%20mol%7D%7B1.35%20mol%2B1.94%20mol%7D%3D0.4103%20)
Similarly, mole fraction of oxygen
![\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897](https://tex.z-dn.net/?f=%5Cchi_2%3D%5Cfrac%7Bn_2%7D%7Bn_1%2Bn_2%7D%3D%5Cfrac%7B1.94%20mol%7D%7B1.35%20mol%2B1.94%20mol%7D%3D0.5897)
Partial pressure of each gas can be calculated by the help of Dalton's' law:
The total pressure is 405 mmHg.
P = 405 mmHg
Partial pressure of nitrogen gas:
![p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg](https://tex.z-dn.net/?f=p_1%3DP%5Ctimes%20%5Cchi_1%3D405%20mmHg%5Ctimes%200.4103%20%3D166.17%20mmHg)
Partial pressure of oxygen gas:
![p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg](https://tex.z-dn.net/?f=p_2%3DP%5Ctimes%20%5Cchi_2%3D405%20mmHg%5Ctimes%200.5897%3D238.83%20mmHg)