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← JEE Advanced 2017 · Paper 1
PhysicsSingle correct+3 / -1Question 17 of 54

JEE Advanced 2017 · Paper 1, Physics question 17

Passage

Answer Q.16, Q.17 and Q.18 by appropriately matching the information given in the three columns of the following table.

An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P−VP-V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. WW 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\gamma is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is nn.

Column 1 Column 2 Column 3
(I) W1→2=1γ−1(P2V2−P1V1)W_{1\to2} = \frac{1}{\gamma - 1}(P_2V_2 - P_1V_1) (i) Isothermal (P) [P-V plot]
(II) W1→2=−PV2+PV1W_{1\to2} = -PV_2 + PV_1 (ii) Isochoric (Q) [P-V plot]
(III) W1→2=0W_{1\to2} = 0 (iii) Isobaric (R) [P-V plot]
(IV) W1→2=−nRTln⁡(V2V1)W_{1\to2} = -nRT\ln\left(\frac{V_2}{V_1}\right) (iv) Adiabatic (S) [P-V plot]
The three-column table for Q.16-18. Column 3 holds four P-V diagrams of a cycle, each highlighting the path from state 1 to state 2: (P) a horizontal arrow at the top of the cycle from 1 to 2 (constant pressure, expansion); (Q) a falling curve from 1 down to 2 along the lower-right side of the cycle (expansion along a curve); (R) a gently falling curve from 1 at the top left to 2 along the top of the cycle (expansion along a curve); (S) a vertical segment between 1 (above) and 2 (below) on the left side of the cycle, at constant volume.

Which one of the following options is the correct combination?

Choose one option
Answer and worked solution

Official answer: B

Option (B), (III)(ii)(S), is consistent throughout:

  • W1→2=0W_{1\to2}=0 (III) means no volume change;
  • that is an isochoric process (ii);
  • on a PP-VV diagram it is the vertical step (S).

Every other option pairs incompatible entries. For example, isochoric with W=−nRTln⁡(V2/V1)≠0W=-nRT\ln(V_2/V_1)\ne0 in (A).

(B)