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NEET
Exam
NEET
Subject
Physics
Chapter
Thermodynamics

NEET · Physics · Thermodynamics

Match the following List - I amp; List - II i) Isothermal process...

Match the following
List - Iamp; List - II
i) Isothermal processamp; a) 0
ii) Isobaric processamp; b) \frac{1}{\gamma-1}\left[P_2 V_2-P_1 V_1\right]
iii) Isochoric processamp; c) \mu RT \ln (V2V1)\left(\frac{V_2}{V_1}\right)
iv) Adiabatic processamp; d) P \left(V_2-V_1\right)
The correct answer is

Options

  1. Ai-c, ii-d, iii-a, iv-b
  2. Bi-a, ii-d, iii-b, iv-c
  3. Ci-c, ii-b, iii-d, iv-a
  4. Di-b, ii-c, iii-a, iv-d

Correct answer

A. i-c, ii-d, iii-a, iv-b

Step-by-step solution

Work done in any quasi-static process (=W=\int_{V_1}^{V_2} P d V). Evaluate this for each process. i) Isothermal process ((T=\mathrm{T}= constant). Use ideal gas: PV=nRTP V=n R T (constant (T \Rightarrow n R T=) constant). (W=\int_{V_1}^{V_2} P d V=\int_{V_1}^{V_2} nRTV\frac{n R T}{V}d V=n R T \ln V2V1\frac{V_2}{V_1} So isothermal (\Rightarrow) match (c). ii) Isobaric process ((P=\mathrm{P}= constant). If pressure is constant, (W=\int_{V_1}^{V_2} P d V=P(V2−V1))Soisobaric(⇒)match(d).iii)Isochoricprocess((\left(V_2-V_1\right)) So isobaric (\Rightarrow) match (d). iii) Isochoric process ((\mathrm{V}=constant).Ifvolumeisconstantconstant). If volume is constantd V=0,therefore(W=∫V1V2PdV=0)Soisochoric(⇒)match(a).iv)Adiabatic(reversible)process., therefore (W=\int_{V_1}^{V_2} P d V=0) So isochoric (\Rightarrow) match (a). iv) Adiabatic (reversible) process. For adiabatic reversible: (P V^\gamma=constant. Let (P V^\gamma=K).Then(W=∫V1V2PdV=K∫V1V2V−γdV=K). Then (W=\int_{V_1}^{V_2} P d V=K \int_{V_1}^{V_2} V^{-\gamma} d V=\frac{K}{1-\gamma}\left(V_2^{1-γ\gamma}-V_1^{1-γ\gamma}\right)) Using (K=P_1 V_1^γ=\gamma=P_2 V_2^γ\gamma and simplifying gives the standard form (W=P1V1−P2V2γ−1.\frac{P_1 V_1-P_2 V_2}{\gamma-1} . So adiabatic (\Rightarrow) match (b). Final matching: i - c, ii −d-d, iii −a-a, iv −b-b.
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