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 T₁$T_1$ to T₂$T_2$ is proportional to (T₂ - T₁)$(T_2 - T_1)$
Choose the correct answer from the options given below:
A.(A), (C), (D) only$\text{(A), (C), (D) only}$
B.(A), (C), (E) only$\text{(A), (C), (E) only}$
C.(B), (E) only$\text{(B), (E) only}$
D.(B), (D) only$\text{(B), (D) only}$
Solution & Explanation
Related Formula
First Law of Thermodynamics:
dQ = dU + dW$dQ = dU + dW$
In an adiabatic process:
dQ = 0 dW = -dU$$dQ = 0 \implies dW = -dU$$
Change in Internal Energy:
dU = n Cv dT = n Cv (T₂ - T₁)$$dU = n C_v dT = n C_v (T_2 - T_1)$$
Core Logic
Let's analyze each statement:
(A) Internal energy is constant: Incorrect. Since temperature changes during an adiabatic expansion/compression, internal energy (U ∝ 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 Vγ-1 = constant$T V^{\gamma-1} = \text{constant}$.
(D) Product of pressure and volume is constant: Incorrect. The relation is P V^γ = constant$P V^\gamma = \text{constant}$.
(E) Work done is proportional to (T₂ - T₁)$(T_2 - T_1)$: Correct. Since dW = -dU = -n Cv (T₂ - T₁)$dW = -dU = -n C_v (T_2 - T_1)$, work done is directly proportional to the temperature change (T₂ - T₁)$(T_2 - T_1)$.
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^γ = C$PV^\gamma = C$, TVγ-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.
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 7
The pressure and volume of an ideal gas are related as PV3/2 = K$PV^{3/2} = K$ (Constant). The work done when the gas is taken from state A (P₁, V₁, T₁$P_1, V_1, T_1$) to state B (P₂, V₂, T₂$P_2, V_2, T_2$) is:
This represents the default standard convention for work executed by the gas medium.
Pattern Recognition
Polytropic model work formula is extremely useful. If x > 1$x > 1$, the denominator 1-x$1-x$ flips the index sequence from (2-1)$(2-1)$ to (1-2)$(1-2)$.
Chapter Mix
Class 11 Physics: Thermodynamics
Q40jee_main_2024_27_jan_morningFirst Law of Thermodynamics
0.08 kg$0.08\text{ kg}$ of air is heated at constant volume through 5°C$5^{\circ}\text{C}$. The specific heat of air at constant volume is 0.17 kcal/kg°C$0.17\text{ kcal/kg}^{\circ}\text{C}$ and J = 4.18 joule/cal$J = 4.18\text{ joule/cal}$. The change in its internal energy is approximately:
A.318 J$318\text{ J}$
B.298 J$298\text{ J}$
C.284 J$284\text{ J}$
D.142 J$142\text{ J}$
Solution
Related Formula
Q = Δ U + W$$Q = \Delta U + W$$
Since the volume is constant, work done W = 0$W = 0$, leading to:
Δ U = Q = m cv Δ T$$\Delta U = Q = m c_v \Delta T$$
Core Logic
Given values:
m = 0.08 kg$m = 0.08\text{ kg}$cv = 0.17 kcal/kg°C = 0.17 × 10³ cal/kg°C$c_v = 0.17\text{ kcal/kg}^{\circ}\text{C} = 0.17 \times 10^3\text{ cal/kg}^{\circ}\text{C}$Δ T = 5°C$\Delta T = 5^{\circ}\text{C}$J = 4.18 J/cal$J = 4.18\text{ J/cal}$
Isochoric processes channel total thermal input strictly into core state configurations, converting raw metric calories directly to standard mechanical Joules via mechanical equivalent factors (J$J$).
Chapter Mix
Class 11 Physics: Thermodynamics
Qjee_main_2024_29_jan_morningWork Done in a Thermodynamic Process
A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. It's volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be:
The graph plots Pressure (Dyne/cm^2) on the y-axis (with coordinates at 4000 and 8000) versus Volume (m^3) on the x-axis (with marks at 3 and 7). State path goes linearly from A to B, then horizontally from B to C.
The graph plots Pressure (Dyne/cm^2) on the y-axis (with coordinates at 4000 and 8000) versus Volume (m^3) on the x-axis (with marks at 3 and 7). State path goes linearly from A to B, then horizontally from B to C.
A.33800 ~J$33800 \mathrm{~J}$
B.2200 ~J$2200 \mathrm{~J}$
C.600 ~J$600 \mathrm{~J}$
D.1200 ~J$1200 \mathrm{~J}$
Solution
Related Formula
Work done (W$W$) in a thermodynamic process corresponds to the area under the curve on a P-V$P-V$ diagram:
W = ∫ P dV$$W = \int P \, dV$$
Core Logic
Let's compute the work done along the distinct steps AB$AB$ and BC$BC$ matching the values shown in the graph:
The graph plots Pressure (Dyne/cm^2) on the y-axis (with coordinates at 4000 and 8000) versus Volume (m^3) on the x-axis (with marks at 3 and 7). State path goes linearly from A to B, then horizontally from B to C.
Step 1: Calculate Work Done in Process AB
Process AB$AB$ is a linear expansion. Fulfilling the area of the trapezoid underneath path AB$AB$:
Since 800 ~J$800 \mathrm{~J}$ is not present in the options, this question was declared a BONUS.
Pattern Recognition
Always be highly vigilant of mixed unit systems in thermodynamic graphics. Here, pressure is in CGS (Dyne/cm²$\text{Dyne/cm}^2$) while volume is in MKS (m³$\text{m}^3$). Converting early keeps you clear of algebraic traps.
Chapter Mix
Class 11 Physics: Thermodynamics
Qjee_main_2024_30_january_eveningIsothermal and Adiabatic Processes
Choose the correct statement for processes A$\mathrm{A}$ & B$\mathrm{B}$ shown in figure.
A Pressure vs Volume (P-V) graph showing two expansion curves starting from the same initial state. Curve B is steeper than Curve A.
A.PVγ = k for process B and PV = k for process A$\mathrm{PV}^{\gamma} = \mathrm{k} \text{ for process } \mathrm{B} \text{ and } \mathrm{PV} = \mathrm{k} \text{ for process } \mathrm{A}$
B.PV = k for process B and A$\mathrm{PV} = \mathrm{k} \text{ for process } \mathrm{B} \text{ and } \mathrm{A}$
C.Pγ - 1Tγ = k for process B and T = k for process A$\frac{\mathrm{P}^{\gamma - 1}}{\mathrm{T}^{\gamma}} = \mathrm{k} \text{ for process } \mathrm{B} \text{ and } \mathrm{T} = \mathrm{k} \text{ for process } \mathrm{A}$
D.TγPγ - 1 = k for process A and PV = k for process B$\frac{\mathrm{T}^{\gamma}}{\mathrm{P}^{\gamma - 1}} = \mathrm{k} \text{ for process } \mathrm{A} \text{ and } \mathrm{PV} = \mathrm{k} \text{ for process } \mathrm{B}$
In a P-V$P-V$ diagram starting from the same initial state and undergoing expansion, the adiabatic curve is steeper than the isothermal curve.
Looking at the graph, Curve (B) is steeper, so it represents an adiabatic process. Curve (A) is less steep, so it represents an isothermal process.
Step 1: Check Options
Process A (Isothermal): PV = k$PV = k$ or T = k$T = k$.
Process B (Adiabatic): PVγ = k$PV^{\gamma} = k$.
Option (1) states: PVγ = k$PV^{\gamma} = k$ for B and PV = k$PV = k$ for A. This is entirely correct.
Note: For an adiabatic process, Tγ P1-γ = k$T^{\gamma} P^{1-\gamma} = k$, which can be written as TγPγ-1 = k$\frac{T^{\gamma}}{P^{\gamma-1}} = k$. Option (3) proposes Pγ-1Tγ = k$\frac{P^{\gamma-1}}{T^{\gamma}} = k$ for B, which is mathematically equivalent, but Option (1) is the standard and most direct NTA choice.
Pattern Recognition
Slope of isothermal = -P/V$= -P/V$. Slope of adiabatic = -γ P/V$= -\gamma P/V$. Since γ > 1$\gamma > 1$, adiabatic is always steeper. In expansions, adiabatic lies below isothermal.
Chapter Mix
Class 11 Physics: Thermodynamics
Q45jee_main_2024_30_jan_morningPolytropic Processes and Heat Capacity
Two thermodynamical process are shown in the figure. The molar heat capacity for process A and B are CA$C_A$ and CB$C_B$. The molar heat capacity at constant pressure and constant volume are represented by CP$C_P$ and CV$C_V$, respectively. Choose the correct statement.
A plot of log P versus log V for two processes A and B.
A.CB = ∞ , CA = 0$C_B = \infty , C_A = 0$
B.CA = 0 and CB = ∞$C_A = 0 \text{ and } C_B = \infty$
C.CP > CV > CA = CB$C_P > C_V > C_A = C_B$
D.CA > CP > CV$C_A > C_P > C_V$
Solution
Related Formula
C = Cv + (R)/(1 - n) (for PVⁿ = const)$$C = C_v + \frac{R}{1 - n} \quad (\text{for } PV^n = \text{const})$$If P = k V^x, then n = -x$$\text{If } P = k V^x, \text{ then } n = -x$$
Core Logic
For a generic straight line on a P$\log P$ vs V$\log V$ plot, the slope defines the exponent.
P = x V + c ⇒ P = V^x · e^c ⇒ PV-x = Constant$\log P = x \log V + c \Rightarrow P = V^x \cdot e^c \Rightarrow PV^{-x} = \text{Constant}$.
The molar heat capacity is computed using n = -x$n = -x$.
Step 1: Evaluate Process A
For process A: slope = γ$= \gamma$ (given θ > 1$\tan \theta > 1$).
P = γ V ⇒ PV-γ = Constant (here n = -γ)$$\log P = \gamma \log V \Rightarrow PV^{-\gamma} = \text{Constant} \quad (\text{here } n = -\gamma)$$CA = CV + (R)/(1 - (-γ)) = CV + (R)/(1 + γ) (i)$$C_A = C_V + \frac{R}{1 - (-\gamma)} = C_V + \frac{R}{1 + \gamma} \quad \dots (i)$$
Step 2: Evaluate Process B
For process B: slope = 1$= 1$ (given 45^° = 1$\tan 45^\circ = 1$).
P = 1 · V ⇒ PV⁻¹ = Constant (here n = -1)$$\log P = 1 \cdot \log V \Rightarrow PV^{-1} = \text{Constant} \quad (\text{here } n = -1)$$CB = Cv + (R)/(1 - (-1)) = Cv + (R)/(2) (ii)$$C_B = C_v + \frac{R}{1 - (-1)} = C_v + \frac{R}{2} \quad \dots (ii)$$
Step 3: Analyze Heat Capacities
We also know:
CP = Cv + R (iii)$$C_P = C_v + R \quad \dots (iii)$$
Comparing (i), (ii), and (iii):
Since γ > 1$\gamma > 1$ for process A, (R)/(1+γ) < (R)/(2)$\frac{R}{1+\gamma} < \frac{R}{2}$.
Thus, CP > CB > CA > Cv$C_P > C_B > C_A > C_v$.
Note: Ans. By NTA (1 or 2). Ans. By our answer (Bonus). Since no option fully matches the mathematical derivation CP > CB > CA > Cv$C_P > C_B > C_A > C_v$, this was treated as a bonus or flawed option set. We assign index 0 based on official NTA preliminary keys marking 1 or 2.
Pattern Recognition
Slopes on a log-log PV plot represent the polytropic exponent modifier. Positive slopes imply -n > 0$-n > 0$, pulling heat capacity structurally between Cv$C_v$ and CP$C_P$.
Chapter Mix
Class 11 Physics: Thermodynamics
More Thermodynamics Questions — jee_main_2025_02_april_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
