One mole of a diatomic ideal gas y -1-4 is taken through a cyclic process starting from point A- The process A - B is an adiabatic compression- B - C is an isobaric expansion- C - D is an adibatic expansion and D - A is an isochoric process- The volume ratios are V A - V B -16- V C - V B -2and temperature T A -27- C- Find the efficiency of the cycle- A- 32-84 -B- 61-40 -C- 28-56 -D- 35-61 -

The correct option is B 61.40 % From the figure, for adiabatic compression process A→ B, T_AV_A^γ 1=T_BV_B^γ 1 From this, we can write that T_B= ( V_AV_B )^γ 1 ...

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Standard XII

Physics

Heat Engines

One mole of a...

Question

One mole of a diatomic ideal gas $(γ = 1.4)$ is taken through a cyclic process starting from point $A$ . The process $A \to B$ is an adiabatic compression, $B \to C$ is an isobaric expansion, $C \to D$ is an adibatic expansion and $D \to A$ is an isochoric process. The volume ratios are $\frac{V_{A}}{V_{B}} = 16, \frac{V_{C}}{V_{B}} = 2$ and temperature $T_{A} = 27^{\circ} C$ . Find the efficiency of the cycle.

32.84 %

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61.40 %

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28.56 %

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35.61 %

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Solution

The correct option is B $61.40 %$

From the figure, for adiabatic compression process $A \to B$ ,
$T_{A} V_{A}^{γ - 1} = T_{B} V_{B}^{γ - 1}$
From this, we can write that
$T_{B} = {(\frac{V_{A}}{V_{B}})}^{γ - 1} T_{A} \Rightarrow T_{B} = (16)^{\frac{2}{5}} \times 300 \approx 909 K$
For an isobaric process $B \to C$ ,
$\frac{V_{B}}{T_{B}} = \frac{V_{C}}{T_{C}}$
$\Rightarrow T_{C} = T_{B} (\frac{V_{C}}{V_{B}}) = 909 \times 2 \approx 1818 K$
For adiabatic expansion process $C \to D$
$\frac{V_{A}}{V_{B}} = 16$ and $\frac{V_{C}}{V_{B}} = 2$
$\Rightarrow \frac{V_{A}}{V_{C}} = \frac{V_{D}}{V_{C}} = 8$
Also, $T_{C} V_{C}^{γ - 1} = T_{D} V_{D}^{γ - 1} ∴ T_{D} = T_{C} {(\frac{V_{C}}{V_{D}})}^{γ - 1} = 1818 {(\frac{1}{8})}^{\frac{2}{5}} \approx 791 K$

Heat absorbed from $B \to C$
$Q_{1} = n C_{P} (T_{C} - T_{B}) = n \frac{γ R}{γ - 1} (T_{C} - T_{B}) = \frac{\frac{7 R}{5}}{\frac{2}{5}} (1818 - 909) = \frac{7 R}{2} \times 909 \approx 3182 R J$
For process $D \to A$ , heat released
$Q_{2} = n C_{V} (T_{D} - T_{A}) = n \frac{R}{γ - 1} (T_{D} - T_{A}) = \frac{5 R}{2} (791 - 300) = \frac{5 R}{2} \times 491 \approx 1227.5 R J$

Work done during process $A B$
$W_{A B} = - \frac{n R}{γ - 1} (T_{B} - T_{A}) = - \frac{5 R}{2} (909 - 300) = \frac{- 5 R}{2} \times 609 = - 1522.5 R J$
Similarly,
$W_{B C} = n R (T_{C} - T_{B}) = 1 \times R (1818 - 909) = 909 R J$
$W_{C D} = \frac{- n R}{γ - 1} (T_{C} - T_{D}) = \frac{5 R}{2} \times 1027 = 2567.5 R J$
Therefore,
$W_{n e t} = W_{A B} + W_{B C} + W_{C D} = - 1522.5 R + 909 R + 2567.5 R$
$= 1954 R J$
Thus, the efficiency of cycle $(η) = \frac{W_{n e t}}{Q_{1}} \times 100 = \frac{1954 R}{3182 R} \times 100 \approx 61.4 %$
Hence, option (b) is the correct answer.

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