Identify the characteristics of an adiabatic process in a monoatomic gas.
(A) Internal energy is constant.
(B) Work done in the process is equal to the change in internal energy.
(C) The product of temperature and volume is a constant.
(D) The product of pressure and volume is a constant.
(E) The work done to change the temperature from T_1$T_1$ to T_2$T_2$ is proportional to (T_2 - T_1)$(T_2 - T_1)$
Choose the correct answer from the options given below:
### Related Formula
1. First Law of Thermodynamics:
dQ = dU + dW$dQ = dU + dW$
In an adiabatic process:
dQ = 0 implies dW = -dU$$dQ = 0 \implies dW = -dU$$
2. Change in Internal Energy:
dU = n C_v dT = n C_v (T_2 - T_1)$$dU = n C_v dT = n C_v (T_2 - T_1)$$
### Core Logic
Let's analyze each statement:
- **(A) Internal energy is constant:** Incorrect. Since temperature changes during an adiabatic expansion/compression, internal energy (U propto T$U \propto T$) must change.
- **(B) Work done is equal to the change in internal energy:** Correct in magnitude (|dW| = |dU|$|dW| = |dU|$). By definition, dW = -dU$dW = -dU$, which correlates the magnitude of work to the change in internal energy.
- **(C) Product of temperature and volume is constant:** Incorrect. The adiabatic equation of state is T V^gamma-1 = textconstant$T V^{\gamma-1} = \text{constant}$.
- **(D) Product of pressure and volume is constant:** Incorrect. The relation is P V^gamma = textconstant$P V^\gamma = \text{constant}$.
- **(E) Work done is proportional to (T_2 - T_1)$(T_2 - T_1)$:** Correct. Since dW = -dU = -n C_v (T_2 - T_1)$dW = -dU = -n C_v (T_2 - T_1)$, work done is directly proportional to the temperature change (T_2 - T_1)$(T_2 - T_1)$.
### Step 1: Determine the correct option
Since only statements (B) and (E) are correct, the correct option is (3).
### Pattern Recognition
Sees: Characteristics of adiabatic thermodynamic process.
Trap: Confusing adiabatic state relations (PV^gamma = C$PV^\gamma = C$, TV^gamma-1 = C$TV^{\gamma-1} = C$) with isothermal state relations (PV = C$PV = C$, T = C$T = C$).
Shortcut: First law of thermodynamics under dQ=0$dQ=0$ strictly enforces |dW| = |dU|$|dW| = |dU|$, which validates statement B and E immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Keywords:#adiabatic process characteristics work done#JEE Main 2025 Evening Q10#First Law of Thermodynamics adiabatic#monoatomic gas adiabatic temperature change
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_02_april_evening
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