A system undergoes a process where both pressure and volume increase. Which thermodynamic quantity must increase?
Answer: A
If both P and V increase for an ideal gas (PV = nRT), then T must increase, hence internal energy U ∝ T increases. Work done by system depends on process path; entropy and heat depend on specific process.
Q.22Hard
Four moles of an ideal gas are heated from 250 K to 350 K at constant volume. Simultaneously, it is allowed to expand at constant temperature. Which process involves greater entropy change?
Answer: B
Isochoric: ΔS = nCv ln(T_f/T_i) = 4 × Cv × ln(250350). Isothermal: ΔS = nR ln(V_f/V_i). For large expansions, isothermal entropy change is typically larger.
Q.23Hard
Three moles of ideal gas undergo polytropic process with n = 1.5. If temperature increases from 300 K to 450 K, the work done by gas is:
Answer: B
W = nR(T₂-T₁)/(1-n) = 3 × 8.314 × 150/(1-1.5) = 3,741/(-0.5) = -3,741 J (compression), |W| ≈ 3,372 J accounting for polytropic work formula
Q.24Hard
A heat engine operates between 600 K and 300 K reservoirs. It absorbs 5000 J from hot reservoir. For a Carnot engine operating between same temperatures, maximum work output would be:
For a van der Waals gas, the critical point is characterized by:
Answer: A
At critical point, both first and second derivatives of pressure with respect to volume are zero, marking the boundary of liquid-gas phase transition
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Q.26Hard
A reversible process has entropy change ΔS_sys = -100 J/K. The entropy change of universe is:
Answer: C
For reversible process: ΔS_universe = ΔS_sys + ΔS_surr = 0. Since ΔS_sys = -100, ΔS_surr = +100, making total change zero
Q.27Hard
In a free expansion of ideal gas into vacuum, the entropy change of system is:
Answer: B
Free expansion is irreversible with ΔU = 0 and W = 0, so Q = 0. Volume increases, so S = nR ln(V_f/V_i) > 0
Q.28Hard
A monatomic ideal gas undergoes a cyclic process ABCA where: A→B is isothermal expansion, B→C is isochoric process, C→A is adiabatic compression. If at point A, P = 1 atm, V = 1 L, and T = 300 K, and the volume doubles from A to B, find the heat absorbed during the isothermal process.
Answer: C
For isothermal process of ideal gas: Q = nRT ln(V_f/V_i) = W. n = PV/RT = (101325 × 0.001)/(8.314 × 300) ≈ 0.0405 mol. Q = nRT ln(2) = 0.0405 × 8.314 × 300 × ln(2) ≈ 600 ln(2) J
Q.29Hard
For an ideal gas undergoing a polytropic process (PV^n = constant), the heat capacity is C = C_v + R/(1-n). For which value of n does the polytropic process become adiabatic?
Answer: B
For adiabatic process, Q = 0, so C = 0. This occurs when 1-n approaches infinity, which happens when n = γ. At n = γ, PV^γ = constant (adiabatic relation).
Q.30Hard
A gas sample at 300 K has an entropy of 200 J/K. When heated at constant pressure to 600 K, its entropy becomes:
Which thermodynamic process results in maximum work extraction from an ideal gas expanding from the same initial to final states?
Answer: B
For expansion between the same P-V states, isothermal process produces maximum work because W = nRT ln(V_f/V_i) is maximum when temperature is highest throughout the process.
Q.32Hard
A 2 kg mass of ice at 0°C is mixed with 5 kg of water at 80°C in a thermally insulated container. If latent heat of fusion = 3.36 × 10⁵ J/kg and specific heat of water = 4200 J/kg·K, determine the final state of the system.
Answer: A
Heat available from water cooling from 80°C to 0°C: Q = 5 × 4200 × 80 = 1.68 × 10⁶ J. Heat needed to melt ice: Q = 2 × 3.36 × 10⁵ = 6.72 × 10⁵ J. Since 1.68 × 10⁶ > 6.72 × 10⁵, all ice melts. Remaining heat: 1.68 × 10⁶ - 6.72 × 10⁵ = 1.008 × 10⁶ J raises temperature of 7 kg water: ΔT = 1.008 × 10⁶/(7 × 4200) ≈ 34.3°C → final temp ≈ 34°C
Q.33Hard
A gas sample undergoes a process where pressure decreases linearly with volume: P = P₀ - kV, where k is a constant. For 1 mole of ideal gas at constant temperature, what is the work done when volume changes from V₁ to V₂?
Answer: A
Work done: W = ∫P dV = ∫(P₀ - kV) dV from V₁ to V₂ = [P₀V - kV²/2] from V₁ to V₂ = P₀(V₂ - V₁) - k(V₂² - V₁²)/2
Q.34Hard
A charged soap bubble of radius R has surface charge density σ. The excess pressure inside the bubble due to electrostatic force is:
Answer: A
Electrostatic pressure = ε₀E²/2 at surface. E = σ/ε₀ just outside. Excess pressure p = σ²/(2ε₀)
Q.35Hard
A point charge q is placed at distance r from an infinite grounded conducting plane. The force on the charge is:
Answer: A
By method of images, image charge -q is at distance r behind plane. Total distance = 2r. F = kq²/(2r)² = q²/(16πε₀r²)
Q.36Hard
The self-energy of a uniformly charged sphere of radius R and total charge Q is:
Answer: A
Self-energy of uniformly charged sphere: U = 3Q²/(20πε₀R) = 3kQ²/(5R)
Q.37Hard
If potential varies as V = 3x² + 4y in a region, the electric field at point (1,2) is:
Answer: A
E = -∇V = -(∂V/∂x i + ∂V/∂y j) = -(6x i + 4j) = -6i - 4j at (1,2)
Q.38Hard
A charged rod of length L with linear charge density λ is placed along the x-axis. The electric field at a point on the perpendicular bisector at distance y from the center is:
Answer: A
By symmetry, perpendicular components cancel. Axial component: E = λ/(2πε₀y) × L/√(L²/4 + y²) = λL/(2πε₀y√(L²/4 + y²))
Q.39Hard
Two point charges Q and -Q are at distance 2d apart. The potential difference between two points on the perpendicular bisector at distances x and 2x from the midpoint is proportional to:
Answer: C
Using potential superposition and the dipole configuration, ΔV = 2kQd(1/x - 21x) for points on the perpendicular bisector.
Q.40Hard
An insulating rod of length L is uniformly charged with total charge Q. The electric potential at a point on the axis at distance x from one end is:
Answer: A
Integrating potential contributions from small elements: V = (kQ/L)ln[(x+L)/x].
