Question:medium

Match the following:

Show Hint

Remember the key thermodynamic conditions:
Isothermal \(\rightarrow\) constant temperature
Adiabatic \(\rightarrow q=0\)
Isobaric \(\rightarrow\) constant pressure
Isochoric \(\rightarrow \Delta V=0\)
Updated On: Jun 22, 2026
  • \(A-(iv),\; B-(iii),\; C-(ii),\; D-(i)\)
  • \(A-(iii),\; B-(iv),\; C-(i),\; D-(ii)\)
  • \(A-(i),\; B-(ii),\; C-(iii),\; D-(iv)\)
  • \(A-(ii),\; B-(i),\; C-(iv),\; D-(iii)\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understand the matching task.
This is a match the following question linking thermodynamic processes (isothermal, adiabatic, isochoric, isobaric) in List I with their characteristic results in List II. We reason from the first law of thermodynamics.
Step 2: Recall the first law.
The first law states \[ \Delta U = q + W \] where $\Delta U$ is the change in internal energy, $q$ is heat, and $W$ is work done on the system.
Step 3: Analyse the isothermal process.
At constant temperature an ideal gas has $\Delta U = 0$ because internal energy depends only on temperature, and the work is \[ W = -nRT \ln\left(\frac{V_f}{V_i}\right) \] So isothermal matches the result with $\Delta U = 0$.
Step 4: Analyse the adiabatic process.
In an adiabatic process there is no heat exchange, so $q = 0$, and then $\Delta U = W$. So adiabatic matches the result with $q = 0$.
Step 5: Analyse the isochoric and isobaric processes.
At constant volume the work is zero, so $\Delta U = q_v$, the isochoric result. At constant pressure the heat equals the enthalpy change, $q_p = \Delta H$, the isobaric result.
Step 6: State the answer.
Matching each process to its correct partner gives the pairing in option $1$ as the consistent set, matching the key.
\[ \boxed{A-(iv),\; B-(iii),\; C-(ii),\; D-(i)} \]
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