Cyclic processes on a PV diagram

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JEE Advanced 2013 Paper 2, Question 20

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

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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.

  1. G \to E
  2. G \to H
  3. F \to H
  4. F \to G
  1. 160 P_0 V_0 \ln 2
  2. 36 P_0 V_0
  3. 24 P_0 V_0
  4. 31 P_0 V_0

Codes:

    \begin{equation*} \begin{matrix} & {\rm P} & {\rm Q} & {\rm R} & {\rm S} \\ {\rm (A)} & 4 & 3 & 2 & 1 \\ {\rm (B)} & 4 & 3 & 1 & 2 \\ {\rm (C)} & 3 & 1 & 2 & 4 \\ {\rm (D)} & 1 & 3 & 2 & 4 \end{matrix} \end{equation*}

Related Article: Thermodynamic processes on an ideal gas

Solution

Along an isotherm PV remains constant whereas PV^{\gamma} is constant for an adiabatic process. Since \gamma>1 we identify the curve F \to G as an isothermal expansion and F \to H as an adiabatic one. The work done by the gas is simply the area under the PV curve. Since F \to G has the largest area under it, it must match the largest value on the second list, 160 P_0 V_0 \ln 2. That means option (A) should be the answer. We can check explicitly using (9) of our note,

(1)   \begin{equation*} |W_{F \to G}| = P_F V_F \ln \left( \frac{P_F}{P_G} \right) = 32 P_0 V_0 \ln \left( \frac{32 P_0}{P_0} \right) = 160 P_0 V_0 \ln 2 . \end{equation*}

Only option (A) has this match up, and therefore option (A) is correct, as expected.

For completeness, we will also calculate the work done in th other processes. First, we need to compute the volume at H on the adiabatic curve. Since P_F V_F^\gamma = P_H V_H^\gamma with \gamma = 5/3 for a monatomic gas,

(2)   \begin{equation*} 32 P_0 V_0^{\frac{5}{3}} = P_0 V_H^{\frac{5}{3}} \implies V_H = 8 V_0 . \end{equation*}

The magnitude of work done along F \to H is given by (16) of our note,

(3)   \begin{equation*} |W_{F \to H}| = \frac{|P_H V_H - P_F V_F|}{\gamma-1} = \frac{|8 P_0 V_0 - 32 P_0 V_0|}{\frac{5}{3}-1} = 36 P_0 V_0 . \end{equation*}

Finally, the work done along the isobaric lines G \to H and G \to E are

(4)   \begin{align*} |W_{G \to H}| &= P_E (V_G-V_H) = P_0 (32 V_0 - 8 V_0) = 24 P_0 V_0 , \\ |W_{G \to E}| &= P_E (V_G-V_E) = P_0 (32 V_0 - V_0) = 31 P_0 V_0 . \end{align*}

Indeed, this confirms that option (A) is correct.

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