Question

Find the charge appearing on each of the three capacitors shown in figure (31-E2).
Figure

Answer 1

Answer:

According to Le Chatelier’s principle, if the pressure on the system is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more.

i)  

The number of moles of reaction products will increase. In the given reaction,

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2 (g)

PCl  

5

​  

(g)⇌PCl  

3

​  

(g)+Cl  

2

​  

(g)

The number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward reaction. As a result, the number of moles of reaction products will increase.

ii) The number of moles of reaction products will decrease. In the given reaction,

CaO (s) + CO_2 (g) \rightleftharpoons CaCO_3 (s)

CaO(s)+CO  

2

​  

(g)⇌CaCO  

3

​  

(s)

The number of moles of gaseous products is less than that of gaseous reactants. Thus, the reaction will proceed in the backward reaction. As a result, the number of moles of reaction products will decrease.

iii) The number of moles of reaction products remains same. In the given reaction,

3Fe (s) + 4H_2O (g) \rightleftharpoons Fe_3O_4 (s) + 4H_2 (g)

3Fe(s)+4H  

2

​  

O(g)⇌Fe  

3

​  

O  

4

​  

(s)+4H  

2

​  

(g)

The number of moles of gaseous products equal to gaseous reactants. Thus, the decreasing of pressure does not affect the equilibrium. As a result, the number of moles of reaction products will remain same.

Explanation:

Answer 2

The charge appearing on each of the three capacitors are 48 μC, 24 μC, & 24 μC

Explanation:

$\begin{array}{l}{\mathrm{C}_{1}=8 \mu \mathrm{F}, \quad \mathrm{C}_{2}=4 \mu \mathrm{F}, \quad \mathrm{C}_{3}=4 \mu \mathrm{F}} \\{\mathrm{Ceq}=\frac{\left(\mathrm{C}_{2}+\mathrm{C}_{3}\right) \times \mathrm{C}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}} \\{=\frac{8 \times 8}{16}=4 \mu \mathrm{F}}\end{array}$

Where C = Capacitance

• From the above figure capacitors B & C are connected in parallel hence total equivalent capacitance is equal to the sum of the individual capacitances.

Where Q = Charge

C = Capacitance

V = Voltage in the circuit

Q1 = 8 $\times$ 6 = 48 μC  

Q2 = 4 $\times$ 6 = 24 μC

Q3 = 4 $\times$ 6 = 24 μC

Hence the charge appearing on each of the three capacitors are 48 μC, 24 μC, & 24 μC respectively.