Solveeit Logo
Login

Question

Question: In a closed system : if the partial pressure C is doubled then partial pressure will be: A. Twice ...

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. 122\dfrac{1}{{2\sqrt 2 }} times, the original pressure
D. 222\sqrt 2 times its original pressure

Explanation

Solution

-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 PSolid=Pliquid=1{P_{Solid}} = {P_{liquid}} = 1

Now from the equation
Kp=(PB)2×(PC)3{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=(PB′)2×(PC′)3{(P{'_B})^2} \times {(P{'_C})^3}
Kp=(PB′)2×(2PC)3{K_p} = {(P{'_B})^2} \times {(2{P_C})^3}
Kp=(PB)2×8(PC)3{K_p} = {({P_B})^2} \times 8{({P_C})^3} ………… (2)

Now dividing equation (1) by equation (2) ,we get
KpKp=(PB)2×(PC)3(PB′)2×8(PC)3\dfrac{{{K_p}}}{{{K_p}}} = \dfrac{{{{({P_B})}^2} \times {{({P_C})}^3}}}{{{{(P{'_B})}^2} \times 8{{({P_C})}^3}}}
(PB′)2=(PB)28{(P{'_B})^2} = \dfrac{{{{({P_B})}^2}}}{8}
(PB′)=(PB)28(P{'_B}) = \sqrt {\dfrac{{{{({P_B})}^2}}}{8}} =(PB)22\dfrac{{({P_B})}}{{2\sqrt 2 }}
(PB′)=(PB)22(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.
PTotal=P1+P2+P3+........+Pn{P_{Total}} = {P_1} + {P_2} + {P_3} + ........ + {P_n}