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JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2023
- Medium
- MCQ
Match List \(I\) with List \(II\) :
| List \(I\) | List \(II\) |
| \(A\) Isothermal Process | \(I\) Work done by the gas decreases internal energy |
| \(B\) Adiabatic Process | \(II\) No change in internal energy |
| \(C\) Isochoric Process | \(III\) The heat absorbed goes partly to increase internal energy and partly to do work |
| \(D\) Isobaric Process | \(IV\) No work is done on or by the gas |
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(A-II, B-I, C-IV, D-III\)
Step-by-step Solution
Detailed explanation
\(\Delta U = nC _{ v } \Delta T\)
For isothermal process \(T\) is constant
So \(\Delta U =0\)
\(A \longrightarrow II\)
Adiabatic process
\(\Delta Q =0\)
\(\Delta Q =\Delta U +\Delta W\)
\(\quad \Delta U =-\Delta W\)
Work done by gas is positive
So \(\Delta U\) is negative
\(B \longrightarrow I\)
For Isochoric process \(\Delta W =0\)
\(C \longrightarrow IV\)
For Isobaric process
\(\Delta W = P \Delta V \neq 0\)
\(\Delta U = nC _{ V } \Delta T \neq 0\)
Heat absorbed goes partly to increase internal energy and partly do work.
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