NEET Physics - MCQ Practice Questions

Charles's Law: At constant pressure, V/T = constant, so V ∝ T

In isochoric (constant volume) process, W = ∫PdV = 0 since dV = 0

Work in isobaric process: W = P(V_f - V_i) = P(2V - V) = PV

In adiabatic compression, Q=0. Work is done ON the gas (W<0), so ΔU = Q - W = -W > 0. Internal energy increases, causing temperature to rise.

In a cyclic process, the system returns to its initial state, so ΔU = 0. By first law: Q = ΔU + W = W.

Carnot efficiency η = 1 - (T_cold/T_hot) = 1 - ($\frac{300}{600}$) = 1 - 0.5 = 0.5 = 50%.

Isobaric process is defined as a process at constant pressure. Prefix 'iso' means constant, and 'baric' refers to pressure.

In isothermal process, T is constant. From ideal gas law PV = nRT, if T is constant, then PV = constant (Boyle's Law).

By energy conservation: Q_in = W + Q_out, so Q_out = Q_in - W = 500 - 200 = 300 J.

Maximum efficiency is Carnot efficiency: η_max = 1 - T_c/T_h = 1 - ($\frac{400}{500}$) = 1 - 0.8 = 0.2 = 20%.

Using first law: ΔU = Q - W, so Q = ΔU + W = -50 + 30 = -20 J. Heat is released by the system.

Heat and work are path functions because their values depend on the path followed during the process. Internal energy, enthalpy, and Gibbs free energy are state functions.

For an isothermal reversible process: W = nRT ln(V₂/V₁) = P₁V₁ ln(V₂/V₁). Since V₂ > V₁, work is positive (work done by the gas).

In an adiabatic process, Q = 0 (no heat transfer). From first law: ΔU = Q - W = 0 - W = -W, so ΔU = -W.

Efficiency η = W/Q_in = (Q_in - Q_out)/Q_in = (500 - 300)/500 = $\frac{200}{500}$ = 0.40 = 40%

By first law of thermodynamics: ΔU = q - w = 500 - 200 = 300 J (where w is work done by the system)

Internal energy depends only on initial and final states, not on the path. Heat and work are path functions.

Carnot efficiency = 1 - (T_cold/T_hot) = 1 - ($\frac{300}{400}$) = 1 - 0.75 = 0.25 = 25%

At equilibrium, ΔG = 0. When ΔG < 0, reaction is spontaneous; when ΔG > 0, reaction is non-spontaneous.

Efficiency η = W/Q_h = 0.4. Work done W = 0.4 × 1000 = 400 J. Heat rejected Q_c = Q_h - W = 1000 - 400 = 600 J