First slide

Work done by gas

Question

Figure shows a cyclic process ABCDBEA. performed on an ideal cycle. lf $If P_{A} = 2 atm, P_{B} = 5 atm and P_{c} = 6 atm . V_{E} - V_{A} = 20 litre$ , find the work done by the gas in the complete, process (l atm. pressure = $1 \times 10^{5}$ Pa)

Moderate

A

2.67 kJ
B

1.33 kJ
C

3.45 kJ
D

4.25 kJ

Solution

The complete cyclic process can be visualized as made up of two cycles, i.e., cycle AEBA (clockwise) and cycle BDCB (counter-clockwise).
Work done by the gas during the cycle AEBA should be positive,
$W_{1} = area of the loop in the P - V diagram$
= $\frac{1}{2} (base) (altitude) = \frac{1}{2} (V_{E} - V_{A}) (P_{B} - P_{A})$
= $\frac{1}{2} (20 \times 10^{- 3} m^{3}) (5 - 2) \times 10^{5} {Nm}^{- 2} = 3 kJ$
Work done by the gas the cycle BDCB should be negative,
$W_{2} = - (area of loop BDCB)$
Now evidently, triangles ABE and BCD are similar, the corresponding angles being equal (i.e.,
$< EBA = < DBC, < EAB = < BCD$
$\frac{(V_{E} - V_{A})}{(P_{B} - P_{A})} = \frac{(V_{C} - V_{D})}{(P_{C} - P_{B})} \Rightarrow \frac{20}{(5 - 2)} = \frac{(V_{C} - V_{D})}{(6 - 5)}$
$\Rightarrow (V_{C} - V_{D}) = \frac{20}{3} litre$
$W_{2} = \frac{1}{2} \times \frac{20}{3} \times 10^{- 3} \times 1 \times 10^{5} = - 0.33 kJ$
$∴ Total work done by the gas$
$∆ W = W_{1} + W_{2} = 3 kJ - 0.33 kJ = 2.67 kJ$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App