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
608.7k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

