The P-V diagram of an engine is shown in the figure below. The temperatures at points 1, 2, 3 and 4 are T1, T2, T3 and T4, respectively. 1→2 and 3→4 are adiabatic processes, and 2→3 and 4→1 are isochoric processes
Identify the correct statement(s).
[γ is the ratio of specific heats Cp (at constant P) and Cv (at constant V)]
Apply the adiabatic and isochoric process relations.
1. Adiabatic Process Relationships:
- For \(1 \to 2\) (adiabatic process): \(T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}.\)
- For \(3 \to 4\) (adiabatic process): \(T_3 V_3^{\gamma - 1} = T_4 V_4^{\gamma - 1}.\)
- For adiabatic processes, pressure and temperature are related as: \(T_1 T_3 = T_2 T_4.\)
Hence, statement (A) is correct.
2. Efficiency of the Engine:
- The efficiency of a heat engine operating on a cycle is given by:
\(\eta = 1 - \frac{Q_{{out}}}{Q_{{in}}}\), where \(Q_{{out}}\) and \(Q_{{in}}\) are the heat rejected and absorbed, respectively.
- For this cycle, the efficiency can also be expressed using the pressure ratio for adiabatic processes as: \(\eta = 1 - \left(\frac{P_1}{P_2}\right)^{\frac{\gamma - 1}{\gamma}}.\)
Hence, statement (B) is correct.
3. Entropy Change:
- For the entire cyclic process, the system returns to its initial state. As entropy is a state function, the total entropy change over a complete cycle is zero: \(\Delta S_{{cycle}} = 0\).
Hence, statement (C) is correct.
4. Temperature Relationship:
- The relationship \(T_1 T_2 = T_3 T_4\) is incorrect as it does not align with the conditions for adiabatic and isochoric processes.
Hence, statement (D) is incorrect.
An amount of ice of mass \( 10^{-3} \) kg and temperature \( -10^\circ C \) is transformed to vapor of temperature \( 110^\circ C \) by applying heat. The total amount of work required for this conversion is,
(Take, specific heat of ice = 2100 J kg\(^{-1}\) K\(^{-1}\),
specific heat of water = 4180 J kg\(^{-1}\) K\(^{-1}\),
specific heat of steam = 1920 J kg\(^{-1}\) K\(^{-1}\),
Latent heat of ice = \( 3.35 \times 10^5 \) J kg\(^{-1}\),
Latent heat of steam = \( 2.25 \times 10^6 \) J kg\(^{-1}\))
Match List-I with List-II.
At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)