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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2020
  • Medium
  • MCQ

Match the thermodynamic processes taking place in a system with the correct conditions. In the table: \(\Delta Q\) is the heat supplied, \(\Delta W\) is the work done and \(\Delta U\) is change in internal energy of the system
Process Condition
\((I)\) Adiabatic \((A)\; \Delta W =0\)
\((II)\) Isothermal \((B)\; \Delta Q=0\)
\((III)\) Isochoric \((C)\; \Delta U \neq 0, \Delta W \neq 0 \Delta Q \neq 0\)
\((IV)\) Isobaric \((D)\; \Delta U =0\)

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Verified Solution

Answer & Solution

Correct Answer

(A) \(I-B, II-D, III-A, IV-C\)

Step-by-step Solution

Detailed explanation

\((I)\) Adiabatic process \(\Rightarrow \Delta Q=0\) No exchange of heat takes place with surroundings

\((II)\) Isothermal proess \(\Rightarrow\) Temperature remains constant \((\Delta T =0)\)

\(\Delta u =\frac{ F }{2} nR \Delta T \Rightarrow \Delta u =0\)

No change in internal energy \([\Delta u =0]\)

\((III)\) Isochoric process Volume remains constant

\(\Delta V =0\)

\(W =\int P \cdot d V =0\)

Hence work done is zero.

\((IV)\) Isobaric process \(\Rightarrow\) Pressure remains constant

\(W = P . \Delta V \neq 0\)

\(\Delta u =\frac{ F }{2} nR \Delta T =\frac{ F }{2}[ P \Delta V ] \neq 0\)

\(\Delta Q = n C _{ p } \Delta T \neq 0\)

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