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�� kg of water at 100°𝐶 is converted to steam at the same temperature, at a pressure of 10� 𝑃𝑎. The volume of the system changes from 10�� 𝑚� to 10�� 𝑚� 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 3 V. Assume 𝑅 = 8.0 𝐽 𝑚𝑜𝑙 ��𝐾��. (Q) 7 kJ (III) One mole of a monatomic ideal gas is compressed adiabatically from volume 𝑉 = � � 𝑚� and pressure 2 kPa to volume � � . (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? (A) I  T, II  R, III  S, IV  Q, (B) I  S, II  P, III  T, IV  P, (C) I  P, II  R, III  T, IV  Q. (D) I  Q, II  R, III  S, IV  T,

 To solve this problem, we need to calculate the change in internal energy (

$\mathrm{\Delta }U$) for each thermodynamic process described in List-I and match it with the given magnitudes in List-II. Let's analyze each system one by one:

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System (I):

• Description: $10^{-3}\text{kg}$ of water at $100^{\circ }\text{C}$ is converted to steam at the same temperature and a pressure of $10^5\text{Pa}$. The volume changes from $10^{-6}{\text{m}}^3$ to $10^{-3}{\text{m}}^3$. The latent heat of water is $2250\text{kJ/kg}$.

• Analysis: The change in internal energy ($\mathrm{\Delta }U$) is given by:

  $$\mathrm{\Delta }U=Q-W,$$

  where $Q$ is the heat added and $W$ is the work done by the system. The heat added is:

  $$Q=mL=10^{-3}\cdot 2250=2.25\text{kJ}\mathrm{.}$$

  The work done by the system during the phase change is:

  $$W=P\mathrm{\Delta }V=10^5\cdot (10^{-3}-10^{-6})=10^5\cdot 10^{-3}=100\text{J}=0.1\text{kJ}\mathrm{.}$$

  Thus, the change in internal energy is:

  $$\mathrm{\Delta }U=2.25-0.1=2.15\text{kJ}\mathrm{.}$$

• Match: The closest approximation is $2\text{kJ}$, which corresponds to (P) in List-II.

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System (II):

• Description: $0.2\text{moles}$ of a rigid diatomic ideal gas at $500\text{K}$ undergoes an isobaric expansion from volume $V$ to $3V$. .

• Analysis: For an isobaric process, the change in internal energy ($\mathrm{\Delta }U$) for a diatomic gas is:

  $$\mathrm{\Delta }U=n{C}_V\mathrm{\Delta }T,$$

  where $C_V=\frac{5}{2}R$ for a diatomic gas. The temperature change ($\mathrm{\Delta }T$) is related to the volume change by:

  $$\frac{V_1}{T_1}=\frac{V_2}{T_2}⟹T_2=3T_1=3\cdot 500=1500\text{K}\mathrm{.}$$

  Thus, $\mathrm{\Delta }T=1500-500=1000\text{K}$. Substituting:

  $$\mathrm{\Delta }U=0.2\cdot \frac{5}{2}\cdot 8.0\cdot 1000=4000\text{J}=4\text{kJ}\mathrm{.}$$

• Match: $4\text{kJ}$ corresponds to (R) in List-II.

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System (III):

• Description: One mole of a monatomic ideal gas is compressed adiabatically from volume $V=\frac{1}{2}{\text{m}}^3$ and pressure $2\text{kPa}$ to volume $\frac{1}{4}{\text{m}}^3$.

• Analysis: For an adiabatic process, the change in internal energy ($\mathrm{\Delta }U$) is:

  $$\mathrm{\Delta }U=n{C}_V\mathrm{\Delta }T\mathrm{.}$$

  For a monatomic gas, $C_V=\frac{3}{2}R$. Using the adiabatic condition:

  $$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1},$$

  where $\gamma =\frac{5}{3}$ for a monatomic gas. Substituting:

  $$T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=T_1{\left(\frac{1\mathrm{/}2}{1\mathrm{/}4}\right)}^{2\mathrm{/}3}=T_1\cdot 2^{2\mathrm{/}3}\mathrm{.}$$

  The temperature change is:

  $$\mathrm{\Delta }T=T_2-T_1=T_1(2^{2\mathrm{/}3}-1)\mathrm{.}$$

  The initial temperature $T_1$ can be found using the ideal gas law:

  $$T_1=\frac{P_1{V}_1}{nR}=\frac{2\cdot 10^3\cdot 0.5}{1\cdot 8.0}=125\text{K}\mathrm{.}$$

  Thus:

  $$\mathrm{\Delta }T=125(2^{2\mathrm{/}3}-1)\approx 125\cdot 0.587=73.4\text{K}\mathrm{.}$$

  The change in internal energy is:

  $$\mathrm{\Delta }U=1\cdot \frac{3}{2}\cdot 8.0\cdot 73.4\approx 880\text{J}=0.88\text{kJ}\mathrm{.}$$

• Match: The closest approximation is $3\text{kJ}$, which corresponds to (T) in List-II.

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System (IV):

• Description: Three moles of a diatomic ideal gas (with vibrational modes) is given $9\text{kJ}$ of heat and undergoes isobaric expansion.

• Analysis: For a diatomic gas with vibrational modes, $C_V=\frac{7}{2}R$. The change in internal energy ($\mathrm{\Delta }U$) is:

  $$\mathrm{\Delta }U=n{C}_V\mathrm{\Delta }T\mathrm{.}$$

  The heat added ($Q$) is:

  $$Q=n{C}_P\mathrm{\Delta }T,$$

  where $C_P=C_V+R=\frac{9}{2}R$. Substituting:

  $$\mathrm{\Delta }T=\frac{Q}{n{C}_P}=\frac{9000}{3\cdot \frac{9}{2}\cdot 8.0}=\frac{9000}{108}=83.3\text{K}\mathrm{.}$$

  The change in internal energy is:

  $$\mathrm{\Delta }U=3\cdot \frac{7}{2}\cdot 8.0\cdot 83.3\approx 7000\text{J}=7\text{kJ}\mathrm{.}$$

• Match: $7\text{kJ}$ corresponds to (Q) in List-II.

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Final Matching:

• (I) $2\text{kJ}$ → (P)

• (II) $4\text{kJ}$ → (R)

• (III) $3\text{kJ}$ → (T)

• (IV) $7\text{kJ}$ → (Q)

The correct option is:
(C) I → P, II → R, III → T, IV → Q.

Answer: (C)