In Column-$I$ processes and in Column-$II$ the first law of thermodynamics are given. Match them appropriately:
Column-$I$ Column-$II$
$(a)$ Adiabatic $(i)$ $\Delta Q = \Delta U$
$(b)$ Isothermal $(ii)$ $\Delta Q = \Delta W$
$(iii)$ $\Delta U = -\Delta W$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) The first law of thermodynamics is given by $\Delta Q = \Delta U + \Delta W$.
$(a)$ For an adiabatic process,there is no heat exchange with the surroundings,so $\Delta Q = 0$. Substituting this into the first law: $0 = \Delta U + \Delta W$,which implies $\Delta U = -\Delta W$. Thus,$(a)$ matches $(iii)$.
$(b)$ For an isothermal process,the temperature remains constant,which implies the change in internal energy $\Delta U = 0$. Substituting this into the first law: $\Delta Q = 0 + \Delta W$,which implies $\Delta Q = \Delta W$. Thus,$(b)$ matches $(ii)$.
Therefore,the correct matching is $(a-iii), (b-ii)$.

Explore More

Similar Questions

The internal energy of the air in a room of volume $V$ at temperature $T$, with outside pressure $P$ increasing linearly with time, varies as

What are thermodynamic processes?

$A$ cyclic process $ABCA$ is shown in the $PT$ diagram. When represented on a $PV$ diagram,it would be:

Difficult
View Solution

The slopes of isothermal and adiabatic curves are related as

$A$ real gas within a closed chamber at $27^{\circ} C$ undergoes the cyclic process as shown in the figure. The gas obeys the $PV^3 = RT$ equation for the path $A$ to $B$. The net work done in the complete cycle is (assuming $R = 8 \, J/mol \cdot K$): (in $ \, J$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo