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JEE Advanced 2019 Paper 2 Offline
MCQ (Single Correct Answer)
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In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by T$$\Delta $$X where T is temperature of the system and $$\Delta $$X is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas, $$X = {3 \over 2}R\,\ln \left( {{T \over {{T_A}}}} \right) + R\,\ln \left( {{V \over {{V_A}}}} \right)$$

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 44 English Comprehension
If the process carried out on one mole of monoatomic ideal gas is as shown in the PV-diagram with $${p_0}{V_0} = {1 \over 3}R{T_0}$$, the correct match is,

JEE Advanced 2019 Paper 2 Offline Physics - Heat and Thermodynamics Question 44 English
A
I$$ \to $$S, II $$ \to $$ R, III $$ \to $$ Q, IV $$ \to $$ T
B
I $$ \to $$ Q, II $$ \to $$ R, III $$ \to $$ P, IV $$ \to $$ U
C
I $$ \to $$ Q, II $$ \to $$ S, III $$ \to $$ R, IV $$ \to $$ U
D
I $$ \to $$ Q, II $$ \to $$ R, III $$ \to $$ S, IV $$ \to $$ U
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