Match List I with List II : List I List II A Isothermal Process I...

(A) Change in internal energy is expressed as$∆ U = n{C}_{v} ∆ T$, here, $∆ T$ is change in temperature.

Since, in an isothermal process temperature remains constant, thus, $∆ U = 0$

$A \to II$

(B) In an adiabatic process, heat transfer, $Q = 0$.

So, from first law of thermodynamics,

$Q = ∆ U + W$

$∆ U = - W$

Since, work done by gas is positive, thus, $∆ U$ is negative

$B \to I$

(C) In an isochoric process, volume remains constant, so work done by or on the gas $W = P ∆ V = 0$

$C \to IV$

(D) In an isobaric process, pressure remains constant, so work done$W = P ∆ V \ne 0$ and change in internal energy $∆ U = n{C}_{v} ∆ T \ne 0$

Thus, heat absorbed goes partly to increase internal energy and partly do work.

$D \to III$