
A gaseous mixture was prepared by taking an equal mole of CO and ${N_2}$. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (${N_2}$) in the mixture is:
(A) 0.5 atm
(B) 0.8 atm
(C) 0.9 atm
(D) 1 atm
Answer
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Hint: Since the number of moles of CO and ${N_2}$ are the same in the mixture, use the ideal gas equation to find the values of their respective pressures in variable form. Equate the sum of the pressures of both to 1 (since total pressure is 1). This will lead us to pressure of nitrogen in the mixture.
Complete step by step answer:
-The question says that the gaseous mixture was prepared by taking equal moles of CO and ${N_2}$.
So, ${n_{{N_2}}} = {n_{CO}}$. Let moles of both CO and ${N_2}$ be equal to ‘n’.
-Now we need to find the partial pressures of both CO and ${N_2}$. For this we will use the ideal gas equation.
Ideal gas equation: PV = nRT.
For the solution the temperature is T and volume is V.
For ${N_2}$: ${P_{{N_2}}}V = {n_{{N_2}}}RT$
${P_{{N_2}}}V = nRT$
${P_{{N_2}}} = \dfrac{{nRT}}{V}$ (1)
For CO: ${P_{CO}}V = {n_{CO}}RT$
${P_{CO}}V = nRT$
${P_{CO}} = \dfrac{{nRT}}{V}$ (2)
From (1) and (2) we can say that: ${P_{{N_2}}} = {P_{CO}}$, which means that partial pressure of ${N_2}$ and CO are same. So, let: ${P_{{N_2}}} = {P_{CO}}$ = P.
-Also the question says that the total pressure of the mixture is 1 atm.
So, ${P_{{N_2}}} + {P_{CO}} = 1$
This can also be written as: P + P = 1
2P = 1
P = $\dfrac{1}{2}$ = 0.5 atm
So, we can now tell that the pressure of ${N_2}$ and CO in the mixture will be 0.5 atm each.
So, the correct answer is “Option A”.
Note: We did total pressure of the mixture equal to the sum of pressures of ${N_2}$ and CO because of Dalton's Law of partial pressure. It states that the total pressure of a mixture of ideal gases is equal to the sum of partial pressures of the constituent gases. Partial pressure is the pressure that each gas would exert if it occupied the volume of the mixture alone at the temperature of the mixture.
Complete step by step answer:
-The question says that the gaseous mixture was prepared by taking equal moles of CO and ${N_2}$.
So, ${n_{{N_2}}} = {n_{CO}}$. Let moles of both CO and ${N_2}$ be equal to ‘n’.
-Now we need to find the partial pressures of both CO and ${N_2}$. For this we will use the ideal gas equation.
Ideal gas equation: PV = nRT.
For the solution the temperature is T and volume is V.
For ${N_2}$: ${P_{{N_2}}}V = {n_{{N_2}}}RT$
${P_{{N_2}}}V = nRT$
${P_{{N_2}}} = \dfrac{{nRT}}{V}$ (1)
For CO: ${P_{CO}}V = {n_{CO}}RT$
${P_{CO}}V = nRT$
${P_{CO}} = \dfrac{{nRT}}{V}$ (2)
From (1) and (2) we can say that: ${P_{{N_2}}} = {P_{CO}}$, which means that partial pressure of ${N_2}$ and CO are same. So, let: ${P_{{N_2}}} = {P_{CO}}$ = P.
-Also the question says that the total pressure of the mixture is 1 atm.
So, ${P_{{N_2}}} + {P_{CO}} = 1$
This can also be written as: P + P = 1
2P = 1
P = $\dfrac{1}{2}$ = 0.5 atm
So, we can now tell that the pressure of ${N_2}$ and CO in the mixture will be 0.5 atm each.
So, the correct answer is “Option A”.
Note: We did total pressure of the mixture equal to the sum of pressures of ${N_2}$ and CO because of Dalton's Law of partial pressure. It states that the total pressure of a mixture of ideal gases is equal to the sum of partial pressures of the constituent gases. Partial pressure is the pressure that each gas would exert if it occupied the volume of the mixture alone at the temperature of the mixture.
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