(a) Write down Maxwell's equations in a non-conducting medium with constant permeability and susceptibility (ρ = j = 0). Show that E⃗ and B⃗ each satisfies the wave equation, and find an expression for the wave velocity. Write down the plane wave sol

Question

(a) Write down Maxwell's equations in a non-conducting medium with constant permeability and susceptibility (ρ = j = 0). Show that E⃗ and B⃗ each satisfies the wave equation, and find an expression for the wave velocity. Write down the plane wave solutions for E⃗ and B⃗, and show how E⃗ and B⃗ are related. (15 marks)

(b) (i) One mole of gas obeys van der Waals equation of state. If its molar internal energy is given by u = cT – a/V (in which V is the molar volume, a is one of the constants in the equation of state and c is a constant), calculate the molar heat capacities Cv and Cp. (10 marks)

(ii) A compressor designed to compress air is used instead to compress helium. It is found that the compressor overheats. Explain this effect, assuming that the compression is approximately adiabatic and the starting pressure is same for both the gases. [γHe = 5/3, γAir = 7/5] (10 marks)

(c) A gas of interacting atoms has an equation of state and heat capacity at constant volume given by the expressions

p(T, V) = aT¹/2 + bT³ + cV⁻²

Cᵥ(T, V) = dT¹/2 + eT^2V + fT¹/2

where a through f are constants which are independent of T and V. Find the differential of the internal energy dU(T, V) in terms of dT and dV. (15 marks)

UPSC Answer Check · Accepted Answer

(a) Maxwell's equations in a non-conducting medium with ρ=0, j=0 are:
∇·D = 0,
∇·B = 0,
∇×E = -∂B/∂t,
∇×H = ∂D/∂t.
For a linear isotropic medium with constant permittivity ε and permeability μ, D=εE, B=μH. Then:
∇·E = 0,
∇·B = 0,
∇×E = -∂B/∂t,
∇×B = με ∂E/∂t.

Take curl of ∇×E = -∂B/∂t:
∇×(∇×E) = -∂/∂t (∇×B) = -με ∂²E/∂t².
Using the vector identity ∇×(∇×E) = ∇(∇·E) - ∇²E, and ∇·E=0, we get -∇²E = -με ∂²E/∂t², so:
∇²E = με ∂²E/∂t².
Similarly, take curl of ∇×B = με ∂E/∂t:
∇×(∇×B) = με ∂/∂t (∇×E) = -με ∂²B/∂t².
Using ∇×(∇×B) = -∇²B, we get:
∇²B = με ∂²B/∂t².
Thus E and B each satisfy the wave equation with wave velocity:
v = 1/√(με).
If ε = ε₀(1+χₑ) and μ = μ₀(1+χₘ), then v = c/√((1+χₑ)(1+χₘ)).

Plane wave solutions are:
E = E₀ exp[i(k·r - ωt)],
B = B₀ exp[i(k·r - ωt)].
From ∇·E=0 and ∇·B=0, we get k·E₀=0 and k·B₀=0, so both fields are transverse.
From Faraday's law, ∇×E = -∂B/∂t gives i k×E = iω B, hence k×E = ω B. Therefore:
B = (1/ω) k×E = (1/v) n̂×E,
where n̂ is the unit vector in the direction of k. So E, B, and k are mutually perpendicular, E and B are in phase, and the magnitudes satisfy |E| = v|B|.

(b)(i) The van der Waals equation for one mole is:
(p + a/V²)(V - b) = RT, so p = RT/(V-b) - a/V².
The molar internal energy is u = cT - a/V.
At constant volume, Cᵥ = (∂u/∂T)_V = c.
For Cₚ, use the general thermodynamic relation:
Cₚ - Cᵥ = -T (∂p/∂T)_V² / (∂p/∂V)_T.
Compute:
(∂p/∂T)_V = R/(V-b),
(∂p/∂V)_T = -RT/(V-b)² + 2a/V³.
Then:
Cₚ - c = -T [R/(V-b)]² / [-RT/(V-b)² + 2a/V³]
= -T R²/(V-b)² / [(-RT V³ + 2a(V-b)²)/(V³(V-b)²)]
= T R² V³ / [RT V³ - 2a(V-b)²]
= R² T / [RT - 2a(V-b)²/V³].
Thus:
Cᵥ = c,
Cₚ = c + R² T / [RT - 2a(V-b)²/V³]
or equivalently Cₚ = c + R / [1 - 2a(V-b)²/(R T V³)].
Units: Cᵥ and Cₚ are in J mol⁻¹ K⁻¹.

(b)(ii) For an adiabatic process, T p^((1-γ)/γ) = constant. If the starting pressure is the same and the compressor produces the same pressure ratio, then:
T₂/T₁ = (p₂/p₁)^((γ-1)/γ).
For helium, γ = 5/3, so (γ-1)/γ = (2/3)/(5/3) = 2/5 = 0.400.
For air, γ = 7/5, so (γ-1)/γ = (2/5)/(7/5) = 2/7 ≈ 0.286.
Since 0.400 > 0.286, helium reaches a higher final temperature than air for the same compression. Hence the compressor designed for air overheats when used for helium.
If the compressor is of fixed volume-ratio type, then T₂/T₁ = (V₁/V₂)^(γ-1), and γ-1 is 2/3 for He and 2/5 for air, again giving a larger temperature rise for helium.

(c) The thermodynamic identity for dU is:
dU = Cᵥ dT + [T(∂p/∂T)_V - p] dV.
Given:
p = a√T + bT³ + cV⁻²,
Cᵥ = d√T + eT²V + f√T = (d+f)√T + eT²V.
Compute:
∂p/∂T = a/(2√T) + 3bT².
Then:
T(∂p/∂T)_V - p = T[a/(2√T) + 3bT²] - [a√T + bT³ + cV⁻²]
= (a√T)/2 + 3bT³ - a√T - bT³ - cV⁻²
= -(a√T)/2 + 2bT³ - cV⁻².
Therefore:
dU = [(d+f)√T + eT²V] dT + [-(a√T)/2 + 2bT³ - cV⁻²] dV.
Equivalently, in the original Cᵥ terms:
dU = (d√T + eT²V + f√T) dT + (-(a√T)/2 + 2bT³ - cV⁻²) dV.
Note: for the given p and Cᵥ to describe a single thermodynamic state function U, the exactness condition requires e = 6b and a = 0; otherwise the data are thermodynamically inconsistent. The expression above is the formal differential obtained from the standard identity.

(a) Write down Maxwell's equations in a non-conducting medium with constant permeability and susceptibility (ρ = j = 0). Show…

Model answer

What "Derive" is asking you to do

Structure that answers it

Where marks are lost

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Practice this exact question

More from Physics 2022 Paper I