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 | \((D)\; \Delta U =0\) |
Pick an option to check your answer.
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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