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Heat and Thermodynamics question

2019 · Shift 2 · Q53
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Heat and Thermodynamics question

2019 · Shift 2 · Q53

JEE AdvancedPhysicsHeat and ThermodynamicsMCQ+3 / −1
In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by T Δ\DeltaΔ X where T is temperature of the system and Δ\DeltaΔ X is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas, X=32R ln⁡(TTA)+R ln⁡(VVA)X = {3 \over 2}R\,\ln \left( {{T \over {{T_A}}}} \right) + R\,\ln \left( {{V \over {{V_A}}}} \right)X=23​Rln(TA​T​)+Rln(VA​V​) Here, R is gas constant, V is volume of gas, TA and VA are constants. The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities. JEE Advanced 2019 Paper 2 Offline Physics - Heat and Thermodynamics Question 45 English ComprehensionIf the process on one mole of monatomic ideal gas is as shown in the TV-diagram with P0V0=13RT0{P_0}{V_0} = {1 \over 3}R{T_0}P0​V0​=31​RT0​, the correct match is, JEE Advanced 2019 Paper 2 Offline Physics - Heat and Thermodynamics Question 45 English
  1. A
    I →\to→ P, II →\to→ R, III →\to→ T, IV →\to→ S
  2. B
    I →\to→ P, II →\to→ T, III →\to→ Q, IV →\to→ T
  3. C
    I →\to→ S, II →\to→ T, III →\to→ Q, IV →\to→ U
  4. D
    I →\to→ P, II →\to→ R, III →\to→ T, IV →\to→ P
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