An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P−V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n.
Column 1
Column 2
Column 3
(I) W1→2=γ−11(P2V2−P1V1)
(i) Isothermal
(P) (see structure)
(II) W1→2=−PV2+PV1
(ii) Isochoric
(Q) (see structure)
(III) W1→2=0
(iii) Isobaric
(R) (see structure)
(IV) W1→2=−nRTln(V1V2)
(iv) Adiabatic
(S) (see structure)
Which of the following options is the only correct representation of a process in which ΔU=ΔQ−PΔV ?PQRS
(A)(II) (iv) (R)
(B)(III) (iii) (P)
(C)(II) (iii) (S)
(D)(II) (iii) (P)✓ Correct
Answer: (D)
Solution
The first law of thermodynamics states ΔU=ΔQ−W, where W is the work done by the system. For an isobaric process (constant pressure), W=PΔV, so ΔU=ΔQ−PΔV. This matches the given equation. Looking at the table, (II) gives the work formula for an isobaric process: W1→2=−PV2+PV1=−P(V2−V1)=−PΔV (work done on the system is negative of work done by the system). Column 2 entry (iii) is "Isobaric". Column 3 entry (P) should show an isobaric process (horizontal line on a P-V diagram). Thus, the correct combination is (II) (iii) (P), which is option (D).
JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Single correct option
