The figure shows the plot of an ideal gas taken through a cycle . The part is a semi-circle and is half of an ellipse. Then, [2009]

Question

The figure shows the $P - V$ plot of an ideal gas taken through a cycle $A B C D A$ . The part $A B C$ is a semi-circle and $C D A$ is half of an ellipse. Then, [2009]

Smart Achievers · Accepted Answer

(2, 4)

$(1) Process A \to B is not isothermal. In case of an isothermal$ $process we get a rectangular hyperbola in a P - V diagram.$

$(2) In process B \to C \to D, Δ U is negative. P V decreases and$ $volume also decreases, therefore W is negative.$

$From first law$ $of thermodynamic, Q is negative i.e., there is a heat loss.$

$(3) W_{A B} > W_{B C} .$ $Therefore work done during path A \to B \to C is positive.$

$(4) Work done in clockwise cycle in a P - V diagram is positive.$

Solution

Q.
The figure shows the $P - V$ plot of an ideal gas taken through a cycle $A B C D A$ . The part $A B C$ is a semi-circle and $C D A$ is half of an ellipse. Then, [2009]

1 The process during the path $A \to B$ is isothermal.

2 Heat flows out of the gas during the path $B \to C \to D$ .

3 Work done during the path $A \to B \to C$ is zero.

4 Positive work is done by the gas in the cycle $A B C D A$ .

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Solution

Q. 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, [2009]

1 The process during the path A→B is isothermal.

2 Heat flows out of the gas during the path B→C→D.

3 Work done during the path A→B→C is zero.