List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

List-I	List-II
(I) $$10^{-3}$$ kg of water at 100°C is converted to steam at the same temperature, at a pressure of $$10^5$$ Pa. The volume of the system changes from $$10^{-6}$$ m$$^3$$ to $$10^{-3}$$ m$$^3$$ in the process. Latent heat of water = 2250 kJ/kg.	(P) 2 kJ
(II) 0.2 moles of a rigid diatomic ideal gas with volume $$V$$ at temperature 500 K undergoes an isobaric expansion to volume 3V. Assume $$R = 8.0$$ J mol$$^{-1}$$ K$$^{-1}$$.	(Q) 7 kJ
(III) One mole of a monatomic ideal gas is compressed adiabatically from volume $$V = \frac{1}{3}$$ m$$^3$$ and pressure 2 kPa to volume $$\frac{V}{8}$$.	(R) 4 kJ
(IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given 9 kJ of heat and undergoes isobaric expansion.	(S) 5 kJ
	(T) 3 kJ

Which one of the following options is correct?

Process I - Water ⟶ Steam at 100 °C, constant pressure $$P = 10^{5}\;{\rm Pa}$$

Heat supplied equals the enthalpy change: $$Q = \Delta H = mL$$.
Given mass $$m = 10^{-3}\;{\rm kg}$$ and latent heat $$L = 2250\;{\rm kJ\,kg^{-1}}$$,

$$\Delta H = 10^{-3}\times 2250\;{\rm kJ}=2.25\;{\rm kJ}$$

The work done on the surroundings is $$W = P\Delta V$$ with
$$\Delta V = 10^{-3}-10^{-6}=9.99\times 10^{-4}\;{\rm m^{3}}$$, hence

$$W = 10^{5}\times 9.99\times 10^{-4}\;{\rm J}\approx 1.0\times 10^{2}\;{\rm J}=0.10\;{\rm kJ}$$

By the first law, $$\Delta U = \Delta H - P\Delta V = 2.25-0.10 = 2.15\;{\rm kJ}\approx 2\;{\rm kJ}$$.
Thus I → (P).

Process II - 0.2 mol rigid diatomic gas, isobaric, $$V\;{\rm to}\;3V$$ at $$T_i = 500\;{\rm K}$$

At constant pressure, $$\dfrac{T_f}{T_i} = \dfrac{V_f}{V_i}=3\; \Rightarrow\; T_f = 1500\;{\rm K}$$, so $$\Delta T = 1000\;{\rm K}$$.

For a rigid diatomic gas, $$C_v = \dfrac{5}{2}R = 20\;{\rm J\,mol^{-1}K^{-1}}$$.
Hence

$$\Delta U = nC_v\Delta T = 0.2\times 20\times 1000 = 4000\;{\rm J}=4\;{\rm kJ}$$

Therefore II → (R).

Process III - 1 mol monatomic gas, adiabatic compression from $$V_i=\dfrac13\;{\rm m^{3}}$$, $$P_i = 2\;{\rm kPa}$$ to $$V_f = \dfrac{V_i}{8}$$

Initial temperature: $$T_i=\dfrac{P_iV_i}{R}= \dfrac{2000\times\dfrac13}{8}=83.3\;{\rm K}$$.

For a monatomic gas, $$\gamma=\dfrac53$$ and $$TV^{\gamma-1}={\rm const}$$.
Thus $$T_f = T_i\left(\dfrac{V_i}{V_f}\right)^{\gamma-1}=83.3\times 8^{2/3}=83.3\times 4 = 333\;{\rm K}$$.

$$\Delta T = 333-83.3 = 250\;{\rm K}$$, and $$C_v = \dfrac32 R = 12\;{\rm J\,mol^{-1}K^{-1}}$$, so

$$\Delta U = nC_v\Delta T = 1\times 12\times 250 = 3000\;{\rm J}=3\;{\rm kJ}$$

Hence III → (T).

Process IV - 3 mol vibrating diatomic gas, isobaric; heat supplied $$Q = 9\;{\rm kJ}$$

For a diatomic gas with vibrations active:
$$C_v = \dfrac{7}{2}R,\quad C_p=C_v+R=\dfrac{9}{2}R$$

At constant pressure, $$Q = nC_p\Delta T \Rightarrow \Delta T = \dfrac{Q}{nC_p}$$.
Internal-energy change: $$\Delta U = nC_v\Delta T = \dfrac{C_v}{C_p}\,Q = \dfrac{7/2}{9/2}\,Q = \dfrac79\,Q$$

$$\Delta U = \dfrac79 \times 9\;{\rm kJ}=7\;{\rm kJ}$$

Thus IV → (Q).

Collecting the matches:
I → P, II → R, III → T, IV → Q

Hence the correct option is Option C.