What is displacement current?

 

Displacement current is the current produced by a changing electric field or changing electric flux. It was introduced by James Clerk Maxwell to resolve a logical inconsistency in Ampere's Circuital Law and to prove that electromagnetic waves exist. [1, 2, 3, 4, 5, 6]

The Core Concept: The Capacitor

To understand displacement current, look at a charging parallel plate capacitor. [1, 2]

• Conduction Current (\(I_{c}\)): This is the physical flow of electrons in the connecting wires. [1, 2, 3]
• The Problem: In the empty space between the capacitor plates, there are no physical wires or moving charges. Yet, a magnetic field is detected there. Ampere's law could not explain this missing link because it required actual charge carriers. [1, 2, 3, 4]
• The Solution: Maxwell proposed that the changing electric field (or electric flux) between the plates acts as a "current". He named this the displacement current (\(I_{d}\)). It is exactly equal to the conduction current, ensuring the total current remains continuous across the circuit.

Henry's law constant for CO 2 ​ in water is 1.67×10 ∘ Pa at 298 K . Calculate the quantity of CO 2 ​ in 500 mL , of soda water when packed under 2.5 atm CO 2 ​ pressure at 298 K .

 

Explanation

To calculate the amount of CO${}_2$ dissolved, we use Henry's Law:

$$P=k_H\times x\_C{O}_2$$

where:\

• $P$ = partial pressure of CO${}_2$ above solution \
• $k_H$ = Henry's Law constant (in Pa)
• $x_{C{O}_2}$ = mole fraction of CO${}_2$ in solution

We will:

1. Convert the pressure to Pascal (SI unit).
2. Use Henry's Law to find $x_{C{O}_2}$.
3. Use the definition of mole fraction to calculate moles of CO${}_2$ present in 500 ml of water.

Step-By-Step Solution

Step 1

Convert the pressure from atm to Pa:

$1\text{atm}=1.013\times 10^5\text{Pa}$

So, $$2.5\text{atm}=2.5\times 1.013\times 10^5=2.5325\times 10^5\text{Pa}$$

Step 2

Apply Henry's Law to find $x_{C{O}_2}$:

$$x_{C{O}_2}=\frac{P}{k_H}=\frac{2.5325\times 10^5}{1.67\times 10^8}$$ $$x_{C{O}_2}=1.517\times 10^{-3}$$

Step 3

Let the number of moles of water in 500 mL: (Since soda water is so dilute that it is as good as water, hence we take water as the SOLVENT and hence MOLAR MASS of WATER i.e. 18g is taken for calculation, intead of that of SODA  WATER (soda water is Carbonic Acid H2CO3   Molar mass 62g) dissolved in WATER)

• Density of water $\approx 1\text{g/mL}$
• Mass of water $=500\text{mL}\times 1\text{g/mL}=500\text{g}$
• Molar mass of H${}_2$O $=18\text{g/mol}$
• Moles of H${}_2$O $=\frac{500}{18}=27.78\text{mol}$

Let $n$ = moles of CO${}_2$ dissolved in 500 mL of water.

By mole fraction definition: $$x_{C{O}_2}=\frac{n}{n+27.78}$$

Since $n$ is much less than 27.78, $n+27.78\approx 27.78$ (approximation valid for dilute solutions): $$x_{C{O}_2}\approx \frac{n}{27.78}$$ $$1.517\times 10^{-3}=\frac{n}{27.78}$$ $$n=1.517\times 10^{-3}\times 27.78=0.04214\text{mol}$$

Step 4

Convert moles of CO${}_2$ to mass if required:

Molar mass of CO${}_2$ = $44\text{g/mol}$ $${\text{Mass of CO}}_2=0.04214\times 44=1.85\text{g}$$

Final Answer

The quantity of CO${}_2$ dissolved in 500 mL of soda water at 2.5 atm is:

$$0.0421\text{mol}\text{or}1.85\text{g (rounded to 3 significant figures)}$$