JEE Challenger
More from Thermodynamics
PhyHardAdv 2022 • Paper 1

Thermodynamic Processes Internal Energy Changes Matching List

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-IList-II(I) 10−3 kg of water at 100∘C is converted to steam at the(P) 2 kJsame temperature, at a pressure of 105 Pa. The volume ofthe system changes from 10−6 m3 to 10−3 m3 in the(Q) 7 kJprocess. Latent heat of water =2250 kJ/kg.(R) 4 kJ(II) 0.2 moles of a rigid diatomic ideal gas with volume Vat temperature 500 K undergoes an isobaric expansion to(S) 5 kJvolume 3 V. Assume R=8.0 J mol−1K−1.(T) 3 kJ(III) One mole of a monatomic ideal gas is compressedadiabatically from volume V=13 m3 and pressure 2 kPato volume V8.(IV) Three moles of a diatomic ideal gas whose moleculescan vibrate, is given 9 kJ of heat and undergoes isobaricexpansion.\begin{array}{ll} \text{\textbf{List-I}} & \text{\textbf{List-II}} \\ \text{(I) } 10^{-3}\text{ kg of water at } 100^\circ C\text{ is converted to steam at the} & \text{(P) } 2\text{ kJ} \\ \text{same temperature, at a pressure of } 10^5\text{ Pa. The volume of} & \\ \text{the system changes from } 10^{-6}\text{ m}^3\text{ to } 10^{-3}\text{ m}^3\text{ in the} & \text{(Q) } 7\text{ kJ} \\ \text{process. Latent heat of water } = 2250\text{ kJ/kg.} & \\ & \text{(R) } 4\text{ kJ} \\ \text{(II) } 0.2\text{ moles of a rigid diatomic ideal gas with volume } V & \\ \text{at temperature } 500\text{ K undergoes an isobaric expansion to} & \text{(S) } 5\text{ kJ} \\ \text{volume } 3\text{ V. Assume } R = 8.0\text{ J mol}^{-1}\text{K}^{-1}\text{.} & \\ & \text{(T) } 3\text{ kJ} \\ \text{(III) One mole of a monatomic ideal gas is compressed} & \\ \text{adiabatically from volume } V = \frac{1}{3}\text{ m}^3\text{ and pressure } 2\text{ kPa} & \\ \text{to volume } \frac{V}{8}\text{.} & \\ & \\ \text{(IV) Three moles of a diatomic ideal gas whose molecules} & \\ \text{can vibrate, is given } 9\text{ kJ of heat and undergoes isobaric} & \\ \text{expansion.} & \end{array}

Which one of the following options is correct?

Options

A

I→T,II→R,III→S,IV→Q\text{I} \rightarrow \text{T}, \text{II} \rightarrow \text{R}, \text{III} \rightarrow \text{S}, \text{IV} \rightarrow \text{Q}

B

I→S,II→P,III→T,IV→P\text{I} \rightarrow \text{S}, \text{II} \rightarrow \text{P}, \text{III} \rightarrow \text{T}, \text{IV} \rightarrow \text{P}

C

I→P,II→R,III→T,IV→Q\text{I} \rightarrow \text{P}, \text{II} \rightarrow \text{R}, \text{III} \rightarrow \text{T}, \text{IV} \rightarrow \text{Q}

Correct
D

I→Q,II→R,III→S,IV→T\text{I} \rightarrow \text{Q}, \text{II} \rightarrow \text{R}, \text{III} \rightarrow \text{S}, \text{IV} \rightarrow \text{T}

Topics & Concepts

ThermodynamicsFirst Law of ThermodynamicsKinetic Theory of Gases

Step-by-Step Solution

To determine the correct option, we calculate the magnitude of the change in internal energy (ΔU\Delta U) for each process given in List-I:

Process (I):

  • Given:

    • Mass of water, m=10−3 kgm = 10^{-3} \text{ kg}
    • Initial volume, V1=10−6 m3V_1 = 10^{-6} \text{ m}^3
    • Final volume, V2=10−3 m3V_2 = 10^{-3} \text{ m}^3
    • Pressure, P=105 PaP = 10^5 \text{ Pa}
    • Latent heat of vaporization, L=2250 kJ/kg=2250×103 J/kgL = 2250 \text{ kJ/kg} = 2250 \times 10^3 \text{ J/kg}

  • Calculations:

    • Heat absorbed during phase change: Q=mL=10−3 kg×2250×103 J/kg=2250 JQ = m L = 10^{-3} \text{ kg} \times 2250 \times 10^3 \text{ J/kg} = 2250 \text{ J}
    • Work done by the system during expansion at constant pressure: W=P(V2−V1)=105 Pa×(10−3−10−6) m3≈105×10−3=100 JW = P(V_2 - V_1) = 10^5 \text{ Pa} \times (10^{-3} - 10^{-6}) \text{ m}^3 \approx 10^5 \times 10^{-3} = 100 \text{ J}
    • Using the First Law of Thermodynamics (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) →\rightarrow (P).

Process (II):

  • Given:

    • Number of moles, n=0.2 moln = 0.2 \text{ mol}
    • Gas type: Rigid diatomic gas   ⟹  \implies degrees of freedom f=5f = 5, molar heat capacity Cv=52RC_v = \frac{5}{2} R
    • Initial temperature, T1=500 KT_1 = 500 \text{ K}
    • Gas undergoes isobaric expansion from V1=VV_1 = V to V2=3VV_2 = 3V
    • Universal gas constant, R=8.0 J mol−1K−1R = 8.0 \text{ J mol}^{-1}\text{K}^{-1}

  • Calculations:

    • For an isobaric process, volume is directly proportional to temperature (VT=constant\frac{V}{T} = \text{constant}): T2=T1(V2V1)=500 K×3=1500 KT_2 = T_1 \left(\frac{V_2}{V_1}\right) = 500 \text{ K} \times 3 = 1500 \text{ K} ΔT=T2−T1=1500 K−500 K=1000 K\Delta T = T_2 - T_1 = 1500 \text{ K} - 500 \text{ K} = 1000 \text{ K}
    • The change in internal energy is: ΔU=nCvΔT=n(52R)ΔT\Delta U = n C_v \Delta T = n \left(\frac{5}{2} R\right) \Delta T ΔU=0.2×(52×8.0)×1000=0.2×20×1000=4000 J=4 kJ\Delta U = 0.2 \times \left(\frac{5}{2} \times 8.0\right) \times 1000 = 0.2 \times 20 \times 1000 = 4000 \text{ J} = 4 \text{ kJ}

Thus, (II) →\rightarrow (R).

Process (III):

  • Given:

    • Number of moles, n=1 moln = 1 \text{ mol}
    • Gas type: Monatomic ideal gas   ⟹  f=3\implies f = 3, γ=53\gamma = \frac{5}{3}
    • Initial volume, V1=13 m3V_1 = \frac{1}{3} \text{ m}^3
    • Initial pressure, P1=2 kPa=2000 PaP_1 = 2 \text{ kPa} = 2000 \text{ Pa}
    • Final volume, V2=V18=124 m3V_2 = \frac{V_1}{8} = \frac{1}{24} \text{ m}^3

  • Calculations:

    • For an adiabatic process, P1V1γ=P2V2γP_1 V_1^\gamma = P_2 V_2^\gamma: P2=P1(V1V2)γ=2000×(8)5/3=2000×(23)5/3=2000×32=64000 PaP_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = 2000 \times (8)^{5/3} = 2000 \times (2^3)^{5/3} = 2000 \times 32 = 64000 \text{ Pa}
    • For an adiabatic process, Q=0Q = 0, so ΔU=−W=P2V2−P1V1γ−1\Delta U = -W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1}: P1V1=2000×13=20003 JP_1 V_1 = 2000 \times \frac{1}{3} = \frac{2000}{3} \text{ J} P2V2=64000×124=80003 JP_2 V_2 = 64000 \times \frac{1}{24} = \frac{8000}{3} \text{ J} ΔU=80003−2000353−1=6000323=200023=3000 J=3 kJ\Delta U = \frac{\frac{8000}{3} - \frac{2000}{3}}{\frac{5}{3} - 1} = \frac{\frac{6000}{3}}{\frac{2}{3}} = \frac{2000}{\frac{2}{3}} = 3000 \text{ J} = 3 \text{ kJ}

Thus, (III) →\rightarrow (T).

Process (IV):

  • Given:

    • Number of moles, n=3 moln = 3 \text{ mol}
    • Gas type: Diatomic ideal gas with vibrational modes active: f=5(translational + rotational)+2(vibrational)=7f = 5 (\text{translational + rotational}) + 2 (\text{vibrational}) = 7 Cv=72R,Cp=Cv+R=92RC_v = \frac{7}{2} R, \quad C_p = C_v + R = \frac{9}{2} R
    • Heat supplied during isobaric expansion, Q=9 kJQ = 9 \text{ kJ}

  • Calculations:

    • For an isobaric process, Q=nCpΔTQ = n C_p \Delta T and ΔU=nCvΔT\Delta U = n C_v \Delta T: ΔU=Q×CvCp=9 kJ×72R92R=9 kJ×79=7 kJ\Delta U = Q \times \frac{C_v}{C_p} = 9 \text{ kJ} \times \frac{\frac{7}{2} R}{\frac{9}{2} R} = 9 \text{ kJ} \times \frac{7}{9} = 7 \text{ kJ}

Thus, (IV) →\rightarrow (Q).

Conclusion:

Matching the results:

  • I→P\text{I} \rightarrow \text{P}
  • II→R\text{II} \rightarrow \text{R}
  • III→T\text{III} \rightarrow \text{T}
  • IV→Q\text{IV} \rightarrow \text{Q}

This corresponds to Option C.

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