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Question:

List - I List - II
(I) Work done by the system in process 1→2→3
(II) Change in internal energy in process 1→2→3
(III) Heat absorbed by the system in process 1→2→3
(IV) Heat absorbed by the system in process 1→2
In a thermodynamic process on an ideal monoatomic 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=3/2Rln(T/TA)+Rln(V/VA). 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. Answer the following by appropriately matching the lists based on the information given in the paragraph. If the process on one mole of monatomic ideal gas is as shown in the TV-diagram with P0V0=1/3RT0, the correct match is,

I→P,II→T,III→Q,IV→T

I→S,II→T,III→Q,IV→U

I→P,II→R,III→T,IV→S

I→P,II→R,III→T,IV→P

Solution:

Correct option is D. I→P,II→R,III→T,IV→P
1-2 process is isothermal and 2-3 process is isochoric
(I) W1→2 = nRTln(Vf/Vi) = 1 × RT0ln(3V0/V0) = RT0ln3 = RT0(1.0986) ≈ RT0
W2→3 = (Isochoric process) 0
W1→2→3 = W1→2 + W2→3 = RT0ln3
(I) W1→2→3 = RT0/3ln2
(II) ΔU = f/2nR(Tf - Ti)
ΔU1→2→3 = 3/2[1 × R(T0 - T0/3)] = RT0
(II→R)
(III) Q1→2→3 = ΔU1→2→3 + W1→2→3 = RT0 + RT0/3ln2 = RT0(1 + ln3/3) ≈ RT0(1+0.366) ≈ 1.366RT0
(III→T)
(IV) Q1→2 = ΔU1→2 + W1→2 = 0 + RT0/3ln2 = RT0/3ln2
(IV→P)