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An ideal gas goes through a reversible cycle $a\rightarrow b\rightarrow c\rightarrow d$ has the V - T diagram

shown below. Process $d\rightarrow a\: and \: b\rightarrow c$ are adiabatic .

The corresponding P - V diagram for the process is (all figures are schematic and not drawn to scale) :

Option: 1

Option: 2

Option: 3

Option: 4

Answers (1)

best_answer

As we learnt in

Boyle's Law -

At constant temperature

$PV= constant$

$P_{1}V_{1}= P_{2}V_{2}$

- wherein

For drawing PV diagram

$a\rightarrow b$ & $c\rightarrow d$ is straight line

$\Rightarrow\ \; V\alpha T$

Since PV = nRT

$V\alpha T$ happens when p is constant

$\therefore\ \; a\rightarrow b$ & $c\rightarrow d$ are isobaric.

$d\rightarrow a$ & $b\rightarrow c$ are diabatic.

From slope pressure at $a\rightarrow b$ is larger than pressure at $d\rightarrow c$ .

Correct option is 1.

Posted by

Rishabh

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