
Equal masses of \[S{O_2},C{H_4},\] and \[{O_2}\] are mixed in empty container at \[298K\] , when the total pressure is \[2.1atm\] . The partial pressures of \[C{H_4}\] in the mixture is:
\[0.5atm\]
\[0.75atm\]
\[1.2atm\]
\[0.6atm\]
Answer
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Hint: Given that three gases are mixed in an empty container with equal masses, by taking the mass as \[x\] grams, and determine the mole fraction of methane. By substituting the mole fraction of methane and given total pressure in the below formula gives the partial pressure of methane.
Formula used:
\[{p_{C{H_4}}} = {\chi _{C{H_4}}}.{P_{total}}\]
\[{p_{C{H_4}}}\] is partial pressure of methane
\[{\chi _{C{H_4}}}\] is mole fraction of methane
\[{P_{total}}\] is the total pressure which is given
Complete answer:
Given that the mixture consists of \[S{O_2},C{H_4},\] and \[{O_2}\] with equal masses. Let the mass be \[x\] grams. By taking the mass as \[x\] grams and molar mass of each gas the moles of each gas can be determined.
From the moles, mole fraction of methane gas can be calculated.
As the molar mass of \[S{O_2}\] is \[64gmo{l^{ - 1}}\] , the moles of \[S{O_2}\] will be \[\dfrac{x}{{64}}\]
As the molar mass of \[C{H_4}\] is \[16gmo{l^{ - 1}}\] , the moles of \[C{H_4}\] will be \[\dfrac{x}{{16}}\]
As the molar mass of \[{O_2}\] is \[32gmo{l^{ - 1}}\] , the moles of \[{O_2}\] will be \[\dfrac{x}{{32}}\]
Total number of moles of all gases in a mixture will be \[\dfrac{{7x}}{{64}}\]
The mole fraction of methane is the ratio of moles of methane to total number of moles
\[{\chi _{C{H_4}}} = \dfrac{{\dfrac{x}{{16}}}}{{\dfrac{{7x}}{{64}}}} = \dfrac{4}{7}\]
Substitute the mole fraction of methane, and total pressure in the above formula,
\[{p_{C{H_4}}} = \dfrac{4}{7} \times 2.1 = 1.2atm\]
Thus, partial pressure of methane will be \[1.2atm\]
Note:
According to Rouault’s law, the partial pressure of a gas is equal to the product of mole fraction and total pressure. According to Dalton’s law of partial pressures, the sum of partial pressures in a given mixture is equal to the total pressure. Both these are useful to determine the partial pressure of any gas.
Formula used:
\[{p_{C{H_4}}} = {\chi _{C{H_4}}}.{P_{total}}\]
\[{p_{C{H_4}}}\] is partial pressure of methane
\[{\chi _{C{H_4}}}\] is mole fraction of methane
\[{P_{total}}\] is the total pressure which is given
Complete answer:
Given that the mixture consists of \[S{O_2},C{H_4},\] and \[{O_2}\] with equal masses. Let the mass be \[x\] grams. By taking the mass as \[x\] grams and molar mass of each gas the moles of each gas can be determined.
From the moles, mole fraction of methane gas can be calculated.
As the molar mass of \[S{O_2}\] is \[64gmo{l^{ - 1}}\] , the moles of \[S{O_2}\] will be \[\dfrac{x}{{64}}\]
As the molar mass of \[C{H_4}\] is \[16gmo{l^{ - 1}}\] , the moles of \[C{H_4}\] will be \[\dfrac{x}{{16}}\]
As the molar mass of \[{O_2}\] is \[32gmo{l^{ - 1}}\] , the moles of \[{O_2}\] will be \[\dfrac{x}{{32}}\]
Total number of moles of all gases in a mixture will be \[\dfrac{{7x}}{{64}}\]
The mole fraction of methane is the ratio of moles of methane to total number of moles
\[{\chi _{C{H_4}}} = \dfrac{{\dfrac{x}{{16}}}}{{\dfrac{{7x}}{{64}}}} = \dfrac{4}{7}\]
Substitute the mole fraction of methane, and total pressure in the above formula,
\[{p_{C{H_4}}} = \dfrac{4}{7} \times 2.1 = 1.2atm\]
Thus, partial pressure of methane will be \[1.2atm\]
Note:
According to Rouault’s law, the partial pressure of a gas is equal to the product of mole fraction and total pressure. According to Dalton’s law of partial pressures, the sum of partial pressures in a given mixture is equal to the total pressure. Both these are useful to determine the partial pressure of any gas.
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