### Core Logic
Let's evaluate each process condition based on the first law of thermodynamics:
* **(A) Isothermal:** Continuous constant temperature (Delta T = 0$\Delta T = 0$) implies that the internal energy change of an ideal gas is zero, so Delta U = 0 implies text(IV)$\Delta U = 0 \implies \text{(IV)}$ [cite: 730].
* **(B) Adiabatic:** No thermal energy transfer occurs between the system and surroundings, meaning Delta Q = 0 implies text(II)$\Delta Q = 0 \implies \text{(II)}$ [cite: 731].
* **(C) Isobaric:** Constant pressure process where both volume and temperature typically vary, so internal energy changes continuously, Delta U neq 0 implies text(III)$\Delta U \neq 0 \implies \text{(III)}$ [cite: 732].
* **(D) Isochoric:** Rigid boundary condition at constant volume (Delta V = 0$\Delta V = 0$) ensures work done Delta W = PDelta V = 0 implies text(I)$\Delta W = P\Delta V = 0 \implies \text{(I)}$ [cite: 733].
Putting these together yields: (A)-(IV), (B)-(II), (C)-(III), (D)-(I)[cite: 155, 729].
### Pattern Recognition
Matching 'Isochoric' with zero work done (Delta W=0$\Delta W=0$) or 'Adiabatic' with zero heat exchange (Delta Q=0$\Delta Q=0$) are fundamental definitions that let you rapidly break down multi-choice grids[cite: 731, 733].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Keywords:#isothermal internal energy change#JEE Main 2025 Evening Q17#isochoric process work done zero#adiabatic system heat exchange
More Thermodynamics Previous-Year Questions — Page 6
Q38jee_main_2024_31_jan_morningIsobaric Process
The given figure represents two isobaric processes for the same mass of an ideal gas, then
A Volume vs Temperature (V-T) graph showing two straight lines starting from the origin representing distinct constant pressures P1 and P2.
### Related Formula
PV = nRT$PV = nRT$
### Core Logic
From the Ideal Gas Law:
V = left(fracnRPright) T$$V = \left(\frac{nR}{P}\right) T$$
In a V-T$V-T$ graph, the equation of the line represents y = mx$y = mx$, where the slope m$m$ is:
textSlope = fracnRP$$\text{Slope} = \frac{nR}{P}$$textSlope propto frac1P$$\text{Slope} \propto \frac{1}{P}$$
Thus, a higher slope corresponds to a lower pressure.
### Step 2: Compare Slopes
From the given figure, the slope of line 2 is greater than the slope of line 1:
(textSlope)_2 > (textSlope)_1$$(\text{Slope})_2 > (\text{Slope})_1$$
Therefore, inversely:
P_2 < P_1$P_2 < P_1$
or P_1 > P_2$P_1 > P_2$.
### Pattern Recognition
In V-T graphs, steeper lines mean lower Pressure. In P-T graphs, steeper lines mean lower Volume. It's an inverse inverse slope relationship.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
More Thermodynamics Questions — jee_main_2025_07_april_evening
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