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 | (1) $\Delta$W = 0 |
| (II) Isothermal | (2) $\Delta$Q = 0 |
| (III) Isochoric | (3) $\Delta$U $\ne$ 0, $\Delta$W $\ne$ 0, $\Delta$Q $\ne$ 0 |
| (IV) Isobaric | (4) $\Delta$U = 0 |
Solution
(I) Adiabatic, $\Delta$Q = 0
<br><br>(II) Isothermal, $\Delta$U = 0
<br><br>(III) Isochoric, $\int {pdV}$ = 0 $\Rightarrow$ W = 0
<br><br>(IV) Isobaric process $\Rightarrow$ Pressure remains constant
<br><br>W = P.$\Delta$V $\ne$ 0
<br><br>$\Delta$U $\ne$ 0
<br><br>$\Delta$Q = nC<sub>p</sub>$\Delta$T $\ne$ 0
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
This question is part of PrepWiser's free JEE Main question bank. 271 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.