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

Choose the correct answer from the options given below

 

A. In isothermal process, the temperature remains constant. Hence change in internal energy is zero. (Internal energy $\mathrm{d}U={\mathrm{nC}}_v\mathrm{\Delta T}\text{ here }(\mathrm{\Delta T}=0)$
B. In adiabatic process Q = 0 . Work done on it is $\mathrm{W}=-△\mathrm{U}$. Gas does work and temperature drops so we can say there is decrease in internal energy.
C. In isochoric process, volume is constant. Hence work done by the gas (or) on the gas equal to zero, $\mathrm{W}=\int \mathrm{PdV}\Rightarrow \mathrm{dV}=0\Rightarrow \mathrm{W}=0$
D. In isobaric process pressure is constant. So according to 1^st law of thermodynamics $\mathrm{dQ}=\mathrm{dU}+\mathrm{P}\int \mathrm{dV}$