Q54 / 54 · JEE Advanced 2019 (Paper 2) · Physics

Answer the following by appropriately matching the lists based on the information given in the paragraph.
In a thermodynamics process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by TΔX, where T is temperature of the system and ΔX is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas X=32R lnTTA+R lnVVA. 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.
 

 Column - I Column - II
(a)Work done by the system in process 1→2→3(p)13RT0 ln2
(b)Change in internal energy in process 1→2→3(q)13RT0
(c)Heat absorbed by the system in process 1→2→3(r)RT0
(d)Heat absorbed by the system in process 1→2(s)43RT0
  (t)13RT03+ln 2
  (u)56RT0

If the process on one mole of monatomic ideal gas is as shown in the PV -diagram with P0V0=13RT0, the correct match is
Figure

  • A
    a-s, b-r, c-q, d-t
  • B
    a-q, b-r, c-p, d-u
  • C
    a-q, b-s, c-r, d-u
  • Correct answer
    D
    a-q, b-r, c-s, d-u

Solution

a W1→2→3=W1→2+W2→3
=P02V0-V0+0=P0V0
W1→2→3=P0V0
=RT03a→q
Process 1→2⇒isobaricand 2→3⇒isochoric


b U1→2→3=323P02×2V0-P0V0
=32×2P0V0=3P0V0=RT0b→r


c Q1→2→3=U1→2→3+W1→2→3
=RT0+RT03=4RT03c→s


d Q1→2=nCP∆T
=n52RT2-T1
=52P02V0-P0V0
=52P0V0=52RT03
=56R0T0d→u

Hence, the correct answer is option: 'D'.

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