Solveeit Logo
Login

Question

Question: A mixture contains \( 16 \) grams of oxygen, \( 28 \) grams of nitrogen and \( 8 \) grams of methane...

A mixture contains 1616 grams of oxygen, 2828 grams of nitrogen and 88 grams of methane. Total pressure of the mixture is 740740 mm. What is the partial pressure of nitrogen in mm?
(A) 185
(B) 370
(C) 555
(D) 740

Explanation

Solution

Hint : In order to solve this question we have to use the concept of Dalton's Law of partial pressure. Dalton’s law of partial pressures is a gas law which states that the total pressure which is exerted by a mixture of gases is equal to the sum total of the partial pressures which is exerted by each individual gas in the mixture.

Complete step by step solution:
Partial pressure of a gas can be expressed in terms of mole fraction. Mole fraction of a specific gas in a mixture of gases is said to be equal to the ratio of the partial pressure of that gas to the total pressure exerted by the gaseous mixture. The volume occupied by a specific gas in a mixture can be calculated with this mole fraction.
According to Dalton's Law of partial pressure
pA=χA×Ptotal{p_A} = {\chi _A} \times {P_{total}}
where,
pA={p_A} = partial pressure of gas ‘A’ in the mixture,
χA={\chi _A} = ​ mole fraction of gas A
The formula to find mole fraction of gas A is given as
χA=Number of moles of gas ’A’ (nA)Total number of moles of gases present in the mixture (ntotal){\chi _A} = \dfrac{{Number{\text{ of moles of gas 'A' (}}{{\text{n}}_A})}}{{Total{\text{ number of moles of gases present in the mixture (}}{{\text{n}}_{total}})}}
Ptotal={P_{total}} = ​Total pressure of the mixture of the gases which is given in the question as 740mm740mm
The equation to find the number of moles of gas A is;
nA=mass of gas ′A′ (mA)Molar mass of gas ’A’ (MA) (equation 1){n_A} = \dfrac{{mass{\text{ }}of{\text{ }}gas{\text{ }}'A'{\text{ (}}{{\text{m}}_A})}}{{Molar{\text{ }}mass{\text{ }}of{\text{ }}gas{\text{ 'A' (}}{{\text{M}}_A})}}{\text{ (equation 1)}}
Using, mo2=16g,MO2=32g/mol,    mN2=28g,MN2=28g/mol,mCH4=8g,MCH4=16g/mol{m_{{o_2}}} = 16g,{M_{{O_2}}} = 32g/mol,\;\;{m_{{N_2}}} = 28g,{M_{{N_2}}} = 28g/mol,{m_{C{H_4}}} = 8g,{M_{C{H_4}}} = 16g/mol
Calculating number of moles of each gas by using equation one we get,
nO2=0.5mol,nN2=1mol,nCH4=0.5mol{n_{{O_2}}} = 0.5mol,{n_{{N_2}}} = 1mol,{n_{C{H_4}}} = 0.5mol
Hence the formula to find the Partial pressure of Nitrogen gas is therefore,
pN2=χN2×Ptotal=10.5+1+0.5×740mm=370mm{p_{{N_2}}} = {\chi _{{N_2}}} \times {P_{total}} = \dfrac{1}{{0.5 + 1 + 0.5}} \times 740mm = 370mm
Option (B) is the correct answer.

Note:
The pressure which is exerted by one among the mixture of gasses if it occupies the same volume on its own is called Partial pressure. Every gas can exert a certain pressure in a mixture. The total pressure of a mixture of an ideal gas is the sum total of partial pressures of individual gases in the mixture.