An ideal gas goes through a reversible cycle a → b → c → d has the V - T diagram shown…

An ideal gas goes through a reversible cycle $a\to b\to c\to d$ has the V - T diagram shown below. Process $d\to a and b\to c$ are adiabatic.



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

Solution

Is an adiabatic process

$T{V}^{\gamma -1}=const\Rightarrow V{T}^{\frac{1}{\gamma -1}}=const$

$\Rightarrow$ as T increase V decreases at non-uniform rate

In process $a\to b$ P = constant as $V\propto T$

In process $c\to d$ $P^{\prime }=$ constant s $V\propto T$

But since slope of V - T graph $\propto \frac{1}{P}$

since slope of ab < slope of cd

$\Rightarrow P_{ab}>P_{cd}$

Also in adiabatic process $d\to a$ as T is increasing V in decreasing

$\Rightarrow$ P is increasing, so P - V diagram is as below

Asked in: JEE Main 2015 (10 Apr Online)

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