Question

The figure shows the P-V plot of an ideal gas taken through
a cycle ABCDA. The part ABC is a semi-circle and CDA is
half of an ellipse. Then,
(a) the process during the path A → B is isothermal
(b) heat flows out of the gas during the path B → C → D
(c) work done during the path A → B → C is zero
(d) positive work is done by the gas in the cycle ABCDA

Answer 1

Answer:

(a) Process is not isothermal as for an isothermal process (PV=constant) has a curve of hyperbolic shape, but here the process AB is a part of a circle.

(b) In Process BCD, Volume decreases and temperature decreases ΔU = negative,

So, ΔQ = negative

(c) Work done in process A→B→C cannot be zero as the work done is given by the area under the PV graph which is surely not zero in the process ABC.

(d) Cycle is clockwise, so work done by the gas is positive.

Explanation:

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