
In a closed system : if the partial pressure C is doubled then partial pressure will be:
A. Twice the original pressure
B. Half of its original pressure
C. $\dfrac{1}{{2\sqrt 2 }}$ times, the original pressure
D. $2\sqrt 2 $ times its original pressure
Answer
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Hint: -Calculate the partial pressure exerted by the gas B in the given equation by using the conservation law of pressure and then comparing it to the new partial pressure after partial pressure of C is doubled.
Complete step by step answer:
Since active masses of pure solid and liquid is always 1.The molar concentration is always directly proportional to its density. Since the density of solids and liquids always remains constant , the active mass is therefore taken as 1.
So ${P_{Solid}} = {P_{liquid}} = 1$
Now from the equation
${K_p} = {({P_B})^2} \times {({P_C})^3}$ …………(1)
Now the partial pressure of C PC is doubled then P'C=2PC but Kp remains the same because it depends only on temperature .Now the new partial pressure of B is P'B.
Kp=${(P{'_B})^2} \times {(P{'_C})^3}$
${K_p} = {(P{'_B})^2} \times {(2{P_C})^3}$
${K_p} = {({P_B})^2} \times 8{({P_C})^3}$ ………… (2)
Now dividing equation (1) by equation (2) ,we get
$\dfrac{{{K_p}}}{{{K_p}}} = \dfrac{{{{({P_B})}^2} \times {{({P_C})}^3}}}{{{{(P{'_B})}^2} \times 8{{({P_C})}^3}}}$
${(P{'_B})^2} = \dfrac{{{{({P_B})}^2}}}{8}$
$(P{'_B}) = \sqrt {\dfrac{{{{({P_B})}^2}}}{8}} $=$\dfrac{{({P_B})}}{{2\sqrt 2 }}$
$(P{'_B}) = \dfrac{{({P_B})}}{{2\sqrt 2 }}$
Hence if the partial pressure of C is doubled then the partial pressure of B is reduced by 122 times its original pressure .
So, the correct answer is “Option C”.
Note: The pressure of an individual gas in a mixture of gases is known as its partial pressure.Let us assume that we have a mixture of ideal gases, we can use the ideal gas law or general equation to solve problems involving gases in a mixture.
Dalton's law of partial pressure states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the component gas.
${P_{Total}} = {P_1} + {P_2} + {P_3} + ........ + {P_n}$
Complete step by step answer:
Since active masses of pure solid and liquid is always 1.The molar concentration is always directly proportional to its density. Since the density of solids and liquids always remains constant , the active mass is therefore taken as 1.
So ${P_{Solid}} = {P_{liquid}} = 1$
Now from the equation
${K_p} = {({P_B})^2} \times {({P_C})^3}$ …………(1)
Now the partial pressure of C PC is doubled then P'C=2PC but Kp remains the same because it depends only on temperature .Now the new partial pressure of B is P'B.
Kp=${(P{'_B})^2} \times {(P{'_C})^3}$
${K_p} = {(P{'_B})^2} \times {(2{P_C})^3}$
${K_p} = {({P_B})^2} \times 8{({P_C})^3}$ ………… (2)
Now dividing equation (1) by equation (2) ,we get
$\dfrac{{{K_p}}}{{{K_p}}} = \dfrac{{{{({P_B})}^2} \times {{({P_C})}^3}}}{{{{(P{'_B})}^2} \times 8{{({P_C})}^3}}}$
${(P{'_B})^2} = \dfrac{{{{({P_B})}^2}}}{8}$
$(P{'_B}) = \sqrt {\dfrac{{{{({P_B})}^2}}}{8}} $=$\dfrac{{({P_B})}}{{2\sqrt 2 }}$
$(P{'_B}) = \dfrac{{({P_B})}}{{2\sqrt 2 }}$
Hence if the partial pressure of C is doubled then the partial pressure of B is reduced by 122 times its original pressure .
So, the correct answer is “Option C”.
Note: The pressure of an individual gas in a mixture of gases is known as its partial pressure.Let us assume that we have a mixture of ideal gases, we can use the ideal gas law or general equation to solve problems involving gases in a mixture.
Dalton's law of partial pressure states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the component gas.
${P_{Total}} = {P_1} + {P_2} + {P_3} + ........ + {P_n}$
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