Answer:
1)
[tex]x_{CH_4}=0.412\\x_{O_2}=0.588[/tex]
2)
[tex]n_T=0.161mol[/tex]
3)
[tex]m_{CH_4}=1.06gCH_4\\m_{O_2}=3.03gO_2[/tex]
Explanation:
Hello,
1) In this case, given the partial pressures of both methane and oxygen we can compute the mole fraction by considering each partial pressure as shown below:
[tex]x_{CH_4}=\frac{P_{CH_4}}{P_T}=\frac{0.175atm}{0.175atm+0.250atm}=0.412\\x_{O_2}=\frac{P_{O_2}}{P_T}=\frac{0.250atm}{0.175atm+0.250atm}=0.588[/tex]
2) Now, we use the ideal gas equation to compute the total mass with the total pressure:
[tex]PV=nRT\\\\n_T=\frac{PV}{RT}=\frac{(0.175+0.250)atm*10.5L}{0.082\frac{atm*L}{mol*K}*(65+273.15)K}=0.161mol[/tex]
3) Finally, by using the mole fractions, total moles and molar masses we compute the grams of both methane and oxygen:
[tex]m_{CH_4}=0.412*0.161molCH_4*\frac{16gCH_4}{1molCH_4}=1.06gCH_4\\m_{O_2}=0.588*0.161molO_2*\frac{32gO_2}{1molO_2}=3.03gO_2[/tex]
Best regards.
For the mixture of 0.175 atm CH₄ and 0.250 atm O₂ in a container with a volume of 10.5 L at 65 °C, we have:
1) The mole fraction of CH₄ and O₂ is 0.412 and 0.588, respectively.
2) The total number of moles of gas in the mixture is 0.161 moles.
3) There are 1.06 grams of CH₄ and 3.03 grams of O₂ in the mixture.
1) We calculate the mole fraction of CH₄ and O₂ as follows:
[tex] \chi_{CH_{4}} = \frac{n_{CH_{4}}}{n_{CH_{4}} + n_{O_{2}}} [/tex] (1)
Where n is the number of moles
We can find the number of moles of CH₄ and O₂ with the Ideal gas law:
[tex] PV = nRT [/tex]
Where:
V: is the total volume = 10.5 L
P: is the pressure
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the temperature = 65 °C = 338 K
The number of moles of CH₄ and O₂ is given by:
[tex] n_{CH_{4}} = \frac{P_{CH_{4}}V}{RT} [/tex] (2)
[tex] n_{O_{2}} = \frac{P_{O_{2}}V}{RT} [/tex] (3)
By entering equations (2) and (3) into (1):
[tex] \chi_{CH_{4}} = \frac{\frac{P_{CH_{4}}V}{RT}}{\frac{P_{CH_{4}}V}{RT} + \frac{P_{O_{2}}V}{RT}} [/tex]
Since V, R, and T are the same for CH₄ and O₂, we have:
[tex] \chi_{CH_{4}} = \frac{P_{CH_{4}}}{P_{CH_{4}} + P_{O_{2}}} = \frac{0.175 atm}{0.175 atm + 0.250 atm} = 0.412 [/tex]
Since the sum of the mole fraction of CH₄ and O₂ must be equal to 1, the mole fraction of O₂ is:
[tex] \chi_{O_{2}} = 1 - \chi_{CH_{4}} = 1 - 0.412 = 0.588 [/tex]
Therefore, the mole fraction of CH₄ and O₂ is 0.412 and 0.588, respectively.
2) The total number of moles of gas in the mixture is given by the sum of equations (2) and (3):
[tex] n_{T} = n_{CH_{4}} + n_{O_{2}} [/tex]
[tex] n_{T} = \frac{P_{CH_{4}}V}{RT} + \frac{P_{O_{2}}V}{RT} [/tex]
[tex] n_{T} = \frac{V}{RT}(P_{CH_{4}} + P_{O_{2}}) = \frac{10.5 L}{0.082 L*atm/(K*mol)*338 K}(0.175 atm + 0.250 atm) = 0.161 \:moles [/tex]
Hence, the total number of moles of gas in the mixture is 0.161 moles.
3) We can find the mass of CH₄ and O₂ with the molecular weight, as follows:
[tex]m_{CH_{4}} = n_{CH_{4}}*M_{CH_{4}} = \frac{P_{CH_{4}}VM_{CH_{4}}}{RT} = \frac{0.175 atm*10.5 L*16.04 g/mol}{0.082 L*atm/(K*mol)*338 K} = 1.06 g[/tex]
[tex]m_{O_{2}} = n_{O_{2}}*M_{O_{2}} = \frac{P_{O_{2}}VM_{O_{2}}}{RT} = \frac{0.250 atm*10.5 L*31.998 g/mol}{0.082 L*atm/(K*mol)*338 K} = 3.03 g[/tex]
Therefore, there are 1.06 grams of CH₄ and 3.03 grams of O₂ in the mixture.
Learn more here:
I hope it helps you!
