Reasoning: At low density, gas molecules are far apart, so intermolecular forces and molecular volume become negligible. Therefore, the gas behaves approximately like an ideal gas.

Reasoning: A thermos flask prevents heat exchange with the surroundings. The work done in shaking increases the internal energy and temperature of the milk, so the process is approximately adiabatic.

Reasoning: During fasting, the body does not take in food or fluids, which affects its temperature regulation and metabolic processes. The body primarily relies on stored energy, and while it may not maintain a constant temperature (as it can fluctuate), it does not exchange heat with the environment in the same way as during normal feeding.

Reasoning: In a rigid container, the volume remains constant, so no work is done:
$\mathbf{W =}$ $\mathbf{P\Delta V = 0}$
From the first law: $\mathbf{Q = \Delta U + W}$ $\mathbf{\Longrightarrow}$ Therefore, $\mathbf{\Delta U = Q}$

Reasoning: $\mathbf{W = P\Delta V}$ $\mathbf{\Longrightarrow}$ $\mathbf{\Delta V = 20 - 5 = 15}\mathbf{\ }\mathbf{m}^{\mathbf{3}}$
$\mathbf{W =}\mathbf{(}\mathbf{6 \times}\mathbf{10}^{\mathbf{5}}\mathbf{)}\mathbf{(15) = 9 \times}\mathbf{10}^{\mathbf{6}}\mathbf{\ }\mathbf{J}$

Reasoning: During sublimation, a solid changes directly into gas. Gas molecules have much greater randomness and disorder than solid molecules, so entropy increases.

Reasoning: $\mathbf{\eta = 1 -}$ $\frac{\mathbf{T}_{\mathbf{C}}}{\mathbf{T}_{\mathbf{H}}}$ $\mathbf{\Longrightarrow}$ $\mathbf{\eta = 1 -}$ $\frac{\mathbf{300}}{\mathbf{600}}\mathbf{= 0.5}$
Also, $\mathbf{\eta =}$ $\frac{\mathbf{W}}{\mathbf{Q}_{\mathbf{H}}}$ $\mathbf{\Longrightarrow}$ $\mathbf{0.5 =}$ $\frac{\mathbf{800}}{\mathbf{Q}_{\mathbf{H}}}$ $\mathbf{\Longrightarrow}$ $\mathbf{Q}_{\mathbf{H}}\mathbf{= 1600}\mathbf{\ }\mathbf{J}\mathbf{\ }\mathbf{/}\mathbf{\ cycle}$

Reasoning: Convert temperatures into Kelvin.
For the first engine: $\mathbf{T}_{\mathbf{H}}\mathbf{= 200 + 273 = 473}\mathbf{\ }\mathbf{K}\mathbf{\ }$, $\mathbf{T}_{\mathbf{C}}\mathbf{= 273}\mathbf{\ }\mathbf{K}$
$\mathbf{\eta}_{\mathbf{1}}\mathbf{= 1 -}$ $\frac{\mathbf{273}}{\mathbf{473}}\mathbf{=}\frac{\mathbf{200}}{\mathbf{473}}$
For the second engine: $\mathbf{T}_{\mathbf{H}}\mathbf{= 273}\mathbf{\ }\mathbf{K}\mathbf{\ }$, $\mathbf{T}_{\mathbf{C}}\mathbf{= -}\mathbf{\ }\mathbf{200 + 273 = 73}\mathbf{\ }\mathbf{K}$
$\mathbf{\eta}_{\mathbf{2}}\mathbf{= 1 -}$ $\frac{\mathbf{73}}{\mathbf{273}}\mathbf{=}\frac{\mathbf{200}}{\mathbf{273}}$
Therefore, $\frac{\mathbf{\eta}_{\mathbf{1}}}{\mathbf{\eta}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{200}\mathbf{\ }\mathbf{/}\mathbf{\ }\mathbf{473}}{\mathbf{200}\mathbf{\ }\mathbf{/}\mathbf{\ }\mathbf{273}}\mathbf{=}\frac{\mathbf{273}}{\mathbf{473}}\mathbf{\approx 0.577}$

Reasoning: Work done by a gas is equal to the area under the pressure-volume $\mathbf{(P}\mathbf{-}\mathbf{V)}$ curve. During an isobaric process, pressure remains constant while volume increases, so the gas performs more work than in the other listed processes for the same expansion conditions.
$\mathbf{\Longrightarrow}$ $\mathbf{ }\mathbf{W =}$ $\mathbf{P\Delta V}$
Therefore, maximum work is obtained in an isobaric process.

Reasoning: For an ideal refrigerator: $\mathbf{COP =}$ $\frac{\mathbf{T}_{\mathbf{C}}}{\mathbf{T}_{\mathbf{H}}\mathbf{\ }\mathbf{-}\mathbf{\ }\mathbf{T}_{\mathbf{C}}}$
Taking room temperature as $\mathbf{300\ K}$.
$\mathbf{5 =}$ $\frac{\mathbf{T}_{\mathbf{C}}}{\mathbf{300}\mathbf{\ }\mathbf{-}\mathbf{\ }\mathbf{T}_{\mathbf{C}}}$ $\mathbf{\Longrightarrow}$ $\mathbf{1500 - 5}\mathbf{\ }\mathbf{T}_{\mathbf{C}}\mathbf{=}\mathbf{T}_{\mathbf{C}}$
$\mathbf{6}\mathbf{T}_{\mathbf{C}}\mathbf{= 1500}$ $\mathbf{\Longrightarrow}$ $\mathbf{T}_{\mathbf{C}}\mathbf{= 250}\mathbf{\ }\mathbf{K}$

Reasoning: Taking the original temperatures as $\mathbf{T}_{\mathbf{H}}\mathbf{= 600}\mathbf{\ }\mathbf{K}$ and $\mathbf{T}_{\mathbf{C}}\mathbf{= 300}\mathbf{\ }\mathbf{K}$.
$\mathbf{\eta}_{\mathbf{1}}\mathbf{= 1 -}$ $\frac{\mathbf{300}}{\mathbf{600}}\mathbf{= 0.5}$
When the source temperature is doubled:
$\mathbf{T}_{\mathbf{H}}\mathbf{= 1200}\mathbf{\ }\mathbf{K}$ $\mathbf{\Longrightarrow}$ $\mathbf{\eta}_{\mathbf{2}}\mathbf{= 1 -}$ $\frac{\mathbf{300}}{\mathbf{1200}}\mathbf{= 0.75}$ $\mathbf{\Longrightarrow}$ $\frac{\mathbf{\eta}_{\mathbf{2}}}{\mathbf{\eta}_{\mathbf{1}}}\mathbf{=}\frac{\mathbf{0.75}}{\mathbf{0.5}}\mathbf{= 1.5}$
Therefore, the efficiency becomes $\mathbf{1.5\ times}$.

Reasoning: In thermodynamics, a system is defined as the specific part of the universe being studied, and its surroundings are everything outside that system. Systems are classified based on their ability to exchange matter and energy with their surroundings:
● Open System: Both mass and energy (heat and work) can be transferred. ● Closed System: Energy can be transferred, but mass cannot. ● Isolated System: Neither mass nor energy (heat or work) is transferred to or from the environment.

Reasoning: According to the zeroth law of thermodynamics, objects in thermal equilibrium have the same temperature, so there is no net heat flow between them.

Reasoning: The graph shows an isothermal process $\mathbf{(T = constant)}$. For an ideal gas, internal energy depends only on temperature.
Since $\ \mathbf{\Delta T = 0}$, $\mathbf{\Delta U = 0}$.

Reasoning: Internal energy and gravitational potential energy are state functions. Their changes depend only on the initial and final states, not on the path followed.

Reasoning: The gas constant $\mathbf{R}$, Avogadro’s number $\mathbf{N}_{\mathbf{A}}$ and Boltzmann constant $\mathbf{k}$ are related by $\mathbf{R =}\mathbf{N}_{\mathbf{A}}\mathbf{\ }\mathbf{k}$.
Therefore, $\mathbf{k =}$ $\frac{\mathbf{R}}{\mathbf{N}_{\mathbf{A}}}$

Reasoning: Both spheres have the same size, material and temperature change. Thermal expansion depends on the original dimensions, coefficient of expansion and temperature change, not on whether the sphere is hollow or solid.

Reasoning: Different materials have different coefficients of thermal expansion. Therefore, they expand by different amounts for the same increase in temperature.