NEET Physics - MCQ Practice Questions

For same initial and final states, isothermal process produces maximum work because the gas expands at the highest average pressure. W_isothermal > W_adiabatic for expansion.

Negative ΔH° indicates exothermic reaction. Spontaneity depends on both ΔH and ΔS (ΔG = ΔH - TΔS). Low T reactions with negative ΔH are more likely spontaneous.

For adiabatic process: T·P^(1-γ)/γ = constant, so T_f/T_i = (P_f/P_i)^((γ-1)/γ) = 10^(0.$\frac{4}{1}$.4) = 10^($\frac{2}{7}$) ≈ 10^0.286

In throttling, enthalpy H remains constant (isenthalpic process). For ideal gases, temperature also remains constant, but for real gases, it can change.

For reversible processes on the middle reservoir (300K), entropy change must be zero: Q_h,engine/500 = (Q_h,engine - W_engine)/300 and similar relations apply to refrigerator at 300K.

ΔG = ΔH - TΔS = 100000 - T(200). At low T, TΔS is small, so ΔG ≈ 100000 J > 0, non-spontaneous. Only at high T (T > 500 K) is it spontaneous.

For Carnot engine: η = 1 - Tc/Th = 1 - $\frac{300}{400}$ = 0.25. W = ηQh, so 100 = 0.25 × Qh, Qh = 400 J

# Solution: Isothermal Expansion of Ideal Monatomic Gas

In an isothermal process, temperature remains constant, which has important implications for internal energy and entropy changes.

Step 1: Change in Internal Energy

For any ideal gas, internal energy depends only on temperature; since temperature is constant in an isothermal process, the change in internal energy must be zero.

Alternatively, using the first law of thermodynamics:

For an isothermal process: $Q=W=nRT\ln \left(\frac{V_f}{V_i}\right)$

Therefore:

Step 2: Change in Entropy

Entropy change during an isothermal expansion is calculated using the reversible heat transfer divided by temperature.

Substituting values: $n=5$ mol, $R=8.314$ J/mol·K, $V_f=50$ L, $V_i=10$ L

ΔU = 0 J and ΔS = 67.3 J/K

Answer: (A)