Match List-I with List-II.
Match List-I with List-II.
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- (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
- (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
- (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
The Correct Option is C
Approach Solution - 1
To solve the given matching problem, we need to understand the thermodynamic processes and their corresponding equations.
- Isobaric Process (A): In an isobaric process, the pressure remains constant. The work done \(\Delta W\) is associated with the change in volume. The equation for heat transfer in this process is:
- \(\Delta Q = \Delta U + P \Delta V\)
- Isochoric Process (B): In isochoric (or isovolumetric) processes, the volume remains constant, so no work is done (\(\Delta W = 0\)). The heat added is entirely used for change in internal energy:
- \(\Delta Q = \Delta U\)
- Adiabatic Process (C): In an adiabatic process, no heat is exchanged with the surroundings (\(\Delta Q = 0\)). Any change in internal energy is solely due to work done:
- \(\Delta Q = 0\)
- Isothermal Process (D): In an isothermal process, the temperature remains constant, and as per the first law of thermodynamics:
- \(\Delta Q = \Delta W\)
Hence, the correct answer is (A)-(IV), (B)-(II), (C)-(III), (D)-(I).
Approach Solution -2
The relations for heat exchange and work done during thermodynamic processes are:
- (A) Isobaric: In an isobaric process (constant pressure), the heat supplied to the system is equal to the work done by the system, i.e., \( \Delta Q = \Delta W \).
- (B) Isochoric: In an isochoric process (constant volume), the change in heat is equal to the change in internal energy, i.e., \( \Delta Q = \Delta U \).
- (C) Adiabatic: In an adiabatic process (no heat exchange), \( \Delta Q = 0 \).
- (D) Isothermal: In an isothermal process (constant temperature), the change in heat is equal to the change in internal energy plus the work done by the system, i.e., \( \Delta Q = \Delta U + P \Delta V \).
Thus, the correct answer is (3).
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