Chemistry, asked by ISOLATEDREX6591, 1 year ago

Q. ​An ideal gas is taken around the cycle ABCA as shown in P-V diagram. What is the net work done by the gas during the cycle?
12 P1V1
6 P1V1
5P1V1
P1V1

Answers

Answered by Dhruv4886
5

Net work done by the gas during the cycle is 5 P₁V₁

  • We know that total work done in the process is equals to the work done in the process AB + work done in the process BC + work done in the process CA

Work done in the process AB is isochoric, so ΔV = 0, W = 0

Work done in the process BC is

W BC = - P ext (V₂-V₁)

From the graph it is very clear that P shows linear variation.

P avg BC = 1/2 × (6 P₁+P₁) = 7 P₁/2

W BC = - 7 P₁/2 (3 V₁- V₁) = - 7 P₁V₁

Work done in process CA is isobaric process.

So,

W = -Pext (V₂-V₁)

W CA = -P₁ (V₁-3V₁) = 2 P₁V₁

Now work done in process ABCA is

W ABCA = -7 P₁V₁ + 2 P₁V₁

W ABCA = -5 P₁V₁

Negative sign indicates work is done by the gas.

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