Vardaan Institute - Detailed Solutions: Master Sheet (Thermodynamics)

Section A: First Law Solutions

Find change in internal energy ($\Delta Q = 500\text{ J}, W = 200\text{ J}$).

Sol: $\Delta U = \Delta Q - W = 500\text{ J} - 200\text{ J} = 300\text{ J}$.

Work done derivations for Isothermal and Adiabatic processes.

Sol: Isothermal: $W = \int_{V_1}^{V_2} P dV = n R T \int \frac{dV}{V} = n R T \ln\left(\frac{V_2}{V_1}\right)$. Adiabatic: $P V^\gamma = K \implies W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{n R (T_1 - T_2)}{\gamma - 1}$.

Section B: Heat Engines & Carnot Cycle Solutions

Carnot efficiency between $500\text{ K}$ and $300\text{ K}$.

Sol: $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 1 - 0.6 = 0.40 = 40\%$.

Second Law statements.

Sol: Kelvin-Planck: No engine operating in a cycle can convert all absorbed heat into work without rejecting heat to a colder reservoir. Clausius: Heat cannot spontaneously flow from a colder body to a hotter body without external work.

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