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JEE Advanced 2022 · Thermodynamics

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 kg10^{-3}\,\text{kg} of water at 100∘C100^\circ\text{C} is converted to steam at the same temperature, at a pressure of 105 Pa10^{5}\,\text{Pa}. The volume of the system changes from 10−6 m310^{-6}\,\text{m}^{3} to 10−3 m310^{-3}\,\text{m}^{3} in the process. Latent heat of water =2250 kJ/kg= 2250\,\text{kJ/kg}. | (P) 2 kJ2\,\text{kJ} | | (II) 0.20.2 moles of a rigid diatomic ideal gas with volume VV at temperature 500 K500\,\text{K} undergoes an isobaric expansion to volume 3V3V. Assume R=8.0 J mol−1 K−1R = 8.0\,\text{J}\,\text{mol}^{-1}\,\text{K}^{-1}. | (Q) 7 kJ7\,\text{kJ} | | (III) One mole of a monatomic ideal gas is compressed adiabatically from volume V=13 m3V = \frac{1}{3}\,\text{m}^{3} and pressure 2 kPa2\,\text{kPa} to volume V8\frac{V}{8}. | (R) 4 kJ4\,\text{kJ} | | (IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given 9 kJ9\,\text{kJ} of heat and undergoes isobaric expansion. | (S) 5 kJ5\,\text{kJ} | | | (T) 3 kJ3\,\text{kJ} | Which one of the following options is correct?

Answer: C. (I) →\to (P); (II) →\to (R); (III) →\to (T); (IV) →\to (Q)

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Written solution

Step 1: Analyze Process (I).

Calculate the heat supplied QQ during vaporization.

Q=mLQ = m L Q=10−3 kg×2250×103 J/kg=2250 JQ = 10^{-3}\,\text{kg} \times 2250 \times 10^{3}\,\text{J/kg} = 2250\,\text{J}

Calculate the work done WW during expansion.

W=PΔVW = P \Delta V W=105 Pa×(10−3−10−6) m3≈100 JW = 10^{5}\,\text{Pa} \times (10^{-3} - 10^{-6})\,\text{m}^{3} \approx 100\,\text{J}

Apply the First Law of Thermodynamics to find the change in internal energy ΔU\Delta U.

ΔU=Q−W\Delta U = Q - W ΔU=2250 J−100 J=2150 J=2.15 kJ≈2 kJ\Delta U = 2250\,\text{J} - 100\,\text{J} = 2150\,\text{J} = 2.15\,\text{kJ} \approx 2\,\text{kJ}

Thus, (I) matches with (P).

Step 2: Analyze Process (II).

For a rigid diatomic gas, the molar heat capacity at constant volume is CV=52RC_V = \frac{5}{2}R.

Since the process is isobaric, the volume is directly proportional to the temperature.

V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2} V500=3VT2\frac{V}{500} = \frac{3V}{T_2} T2=1500 KT_2 = 1500\,\text{K}

Calculate the change in temperature.

ΔT=1500 K−500 K=1000 K\Delta T = 1500\,\text{K} - 500\,\text{K} = 1000\,\text{K}

Calculate the change in internal energy.

ΔU=nCVΔT\Delta U = n C_V \Delta T ΔU=0.2×52×8.0×1000=4000 J=4 kJ\Delta U = 0.2 \times \frac{5}{2} \times 8.0 \times 1000 = 4000\,\text{J} = 4\,\text{kJ}

Thus, (II) matches with (R).

Step 3: Analyze Process (III).

For a monatomic gas, the adiabatic exponent is γ=53\gamma = \frac{5}{3}.

Use the adiabatic relation to find the final pressure P2P_2.

P1V1γ=P2V2γP_1 V_1^{\gamma} = P_2 V_2^{\gamma} P2=P1(V1V2)γP_2 = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma} P2=2 kPa×(8)5/3=2×32=64 kPaP_2 = 2\,\text{kPa} \times (8)^{5/3} = 2 \times 32 = 64\,\text{kPa}

Calculate the change in internal energy.

ΔU=P2V2−P1V1γ−1\Delta U = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} ΔU=32(64×103×124−2×103×13)\Delta U = \frac{3}{2} \left(64 \times 10^{3} \times \frac{1}{24} - 2 \times 10^{3} \times \frac{1}{3}\right) ΔU=32(80003−20003)=3000 J=3 kJ\Delta U = \frac{3}{2} \left(\frac{8000}{3} - \frac{2000}{3}\right) = 3000\,\text{J} = 3\,\text{kJ}

Thus, (III) matches with (T).

Step 4: Analyze Process (IV).

For a diatomic gas with active vibrational modes, the degrees of freedom are f=3 (trans)+2 (rot)+2 (vib)=7f = 3 \text{ (trans)} + 2 \text{ (rot)} + 2 \text{ (vib)} = 7.

The heat capacities are:

CV=72RC_V = \frac{7}{2}R CP=92RC_P = \frac{9}{2}R

For an isobaric process, the heat supplied is Q=nCPΔTQ = n C_P \Delta T and the change in internal energy is ΔU=nCVΔT\Delta U = n C_V \Delta T.

ΔU=CVCPQ\Delta U = \frac{C_V}{C_P} Q ΔU=79×9 kJ=7 kJ\Delta U = \frac{7}{9} \times 9\,\text{kJ} = 7\,\text{kJ}

Thus, (IV) matches with (Q).

Step 5: Match the results.

Combining the matches from the previous steps:

(I)→(P),(II)→(R),(III)→(T),(IV)→(Q)\text{(I)} \to \text{(P)}, \quad \text{(II)} \to \text{(R)}, \quad \text{(III)} \to \text{(T)}, \quad \text{(IV)} \to \text{(Q)}

Answer: The correct matching is (I) →\to (P), (II) →\to (R), (III) →\to (T), (IV) →\to (Q).

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