JEE Advanced 2013 · Paper 2 · Physics Q20

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One mole of a monatomic ideal gas is taken along two cyclic processes E→F→G→E and E→F→H→E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.

P-V diagram: E at (V₀, P₀); E→F vertical up to F at 32P₀; from F a steep curve (adiabatic) down to H and a gentler curve (isotherm) down to G, both at pressure P₀; the horizontal line at P₀ runs back from G through H to E.

List I

P
G→E
Q
G→H
R
F→H
S
F→G

List II

1
$160P_0V_0\ln2$
2
$36P_0V_0$
3
$24P_0V_0$
4
$31P_0V_0$

P → 4; Q → 3; R → 2; S → 1

P → 4; Q → 3; R → 1; S → 2

P → 3; Q → 1; R → 2; S → 4

P → 1; Q → 3; R → 2; S → 4

Answer and worked solution

Official answer: A

States. E $(V_0,P_0)$ , F $(V_0,32P_0)$ . F→G is the isotherm (gentler curve): $V_G=32V_0$ . F→H is the adiabat ( $\gamma=\frac53$ ): $32V_0^{5/3}=V_H^{5/3}$ , so $V_H=8V_0$ .

G→E (isobaric): $P_0(32-1)V_0=31P_0V_0$ → 4
G→H (isobaric): $P_0(32-8)V_0=24P_0V_0$ → 3
F→H (adiabatic): $\dfrac{32P_0V_0-8P_0V_0}{2/3}=36P_0V_0$ → 2
F→G (isothermal): $32P_0V_0\ln32=160P_0V_0\ln2$ → 1

Code 4 3 2 1.

(A)

Topics:Work done by a gas Thermodynamic processes

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