A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine E₁$\mathrm{E}_1$ works between 473K to 373K and engine E₂$\mathrm{E}_2$ works between 373K to 273K. If η₁₂$\eta_{12}$ , η₁$\eta_1$ and η₂$\eta_2$ are the efficiencies of the engines E, E₁$\mathrm{E}_1$ and E₂$\mathrm{E}_2$ , respectively, then
A.η₁₂ < η₁ + η₂$\eta_{12} < \eta_1 + \eta_2$
B.η₁₂ = η₁η₂$\eta_{12} = \eta_1\eta_2$
C.η₁₂ = η₁ + η₂$\eta_{12} = \eta_{1} + \eta_{2}$
D.η₁₂ ≥ η₁ + η₂$\eta_{12} \geq \eta_1 + \eta_2$
Solution & Explanation
Related Formula
η = 1 - TLTH$$\eta = 1 - \frac{\mathrm{T}_L}{\mathrm{T}_H}$$
Core Logic
Let's compute the efficiency parameters explicitly:
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_28_jan_morning
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.
