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Thermodynamic Processes

3 moles of an...

Question

$3$ moles of an ideal monoatomic gas perfoms a cycle shown in the figure. The gas temperature at $A, B, C, D$ are $T_{A} = 400 K,$ $T_{B} = 800 K, T_{C} = 2400 K, T_{D} = 1200 K .$ Find the work done by the gas.

12 kJ

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20 kJ

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120 kJ

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24 kJ

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Solution

The correct option is B $20 kJ$
As process $A B, C D$ pass through origin on $P$ vs $T$ graphs, we can say that $P \propto T \Rightarrow V = Constant$ .
Hence, Isochoric.
$W_{A B} = W_{C D} = 0$

While process $B C, D A$ are isobaric in nature

$W_{B C} = 3 R (T_{C} - T_{B})$
$W_{D A} = 3 R (T_{A} - T_{D})$
$∴$ Total work done
$W_{A B} + W_{C D} + W_{B C} + W_{D A} = 3 R (T_{A} + T_{C} - T_{B} - T_{D})$
$= 3 R (400 + 2400 - 800 - 1200)$
$= 2400 R = 20 kJ$

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Similar questions

3

A, B, C, D

T_{A} = 400 K,

T_{B} = 800 K, T_{C} = 2400 K, T_{D} = 1200 K .

T_{A} = 400 K, T_{B} = 800 K, T_{C} = 2400 K, T_{D} = 1200 K .

T_{A} = 600 K, T_{B} = 800 K, T_{C} = 2200 K

T_{D} = 1200 K

R = \frac{25}{3} J/mole. K

T_{A} = 600 K

T_{B} = 800 K

T_{C} = 2200 K

T_{D} = 1200 K

3

A B C D A

T_{A} = 400 K, T_{B} = 800 K, T_{C} = 2400 K

T_{D} = 1200 K

(R = 8.314 J / m o l e K)

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