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1 mole of an ideal gas undergoes a cyclic change ABCD whwere the (p,v) cordinates are A(5,1) ,B(5,3),C(2,3)AND D(2,1).Pis in atmosphere and V is in litre.calculate the work done along AB,CD BC and DA and also the net work done in the process .GIVEN...i atm=1.01x10

IM SOORY PPL I CANT SEND THE DIGRAM BECOS ITS SAYING ITS INVALID

Hrishant Goswami , 12 Years ago
Grade 10
anser 1 Answers
Askiitians Tutor Team

To calculate the work done during the cyclic process ABCD for the ideal gas, we need to analyze each segment of the cycle based on the coordinates provided. The coordinates represent pressure (p) in atmospheres and volume (V) in liters. The points are as follows: A(5,1), B(5,3), C(2,3), and D(2,1). Let's break down the work done along each segment of the cycle.

Understanding Work Done in Thermodynamics

In thermodynamics, the work done by or on a gas during a process can be calculated using the formula:

W = ∫ PdV

For processes where pressure is constant, the work done can be simplified to:

W = P(Vf - Vi)

Where Vf is the final volume and Vi is the initial volume. For processes where volume is constant, the work done is zero since there is no change in volume.

Calculating Work Done Along Each Segment

Segment AB

From point A to point B, the pressure remains constant at 5 atm while the volume changes from 1 L to 3 L.

Using the formula for work:

WAB = P(Vf - Vi) = 5 \text{ atm} \times (3 \text{ L} - 1 \text{ L}) = 5 \text{ atm} \times 2 \text{ L} = 10 \text{ atm·L}

Segment BC

From point B to point C, the volume remains constant at 3 L while the pressure changes from 5 atm to 2 atm. Since the volume does not change, the work done is:

WBC = 0

Segment CD

From point C to point D, the pressure remains constant at 2 atm while the volume changes from 3 L to 1 L.

Calculating the work done:

WCD = P(Vf - Vi) = 2 \text{ atm} \times (1 \text{ L} - 3 \text{ L}) = 2 \text{ atm} \times (-2 \text{ L}) = -4 \text{ atm·L}

Segment DA

From point D to point A, the volume remains constant at 1 L while the pressure changes from 2 atm to 5 atm. Again, since the volume does not change, the work done is:

WDA = 0

Summing Up the Work Done

Now, let's calculate the net work done over the entire cycle ABCD:

Wnet = WAB + WBC + WCD + WDA

Wnet = 10 \text{ atm·L} + 0 + (-4 \text{ atm·L}) + 0 = 6 \text{ atm·L}

Final Thoughts

The net work done in the cyclic process ABCD is 6 atm·L. This positive value indicates that work is done by the gas during the cycle. Remember that in thermodynamic processes, the area enclosed by the path on a PV diagram also represents the net work done, which can be a helpful visualization for understanding these calculations.

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