Question

One mole of a monatomic gas is taken through a cycle ABCDA as shown in the P-V diagram. Column II give the characteristics involved in the cycle. Match them with each of the processes given in Column I.

Answer 1

(A) In process $A\to B,$ ${}^{\prime }{v}^{\prime }$ is decreasing $\Rightarrow$ W done by gas $=-ve$

Work is done on the gas (5)

At const pressure, $v$ is decreasing $\Rightarrow T$ is also decreasing and $△u=\frac{f}{2}nR△T$

$\Rightarrow$ Internal energy is also decreasing (1)

By First law of thermodynamics -

$dQ=dv+dw$

( dv = -ve , dw = -ve )

So, $dQ=-ve$ $\Rightarrow$ heat is lost (3).

(B) In process $B\to C,$ ${}^{\prime }{v}^{\prime }$ is constant $\Rightarrow dw=0$

At constant ${}^{\prime }{v}^{\prime },$ $p$ is decreasing ${}^{\prime }{T}^{\prime }$ is also decreasing and so, ${}^{\prime }{u}^{\prime }$ is also decreasing (1)

By first law of thermodynamics,

$dQ=-ve$ $\Rightarrow$ heat is lost (3).

(C) In process $C\to D,$ at const process, ${}^{\prime }{v}^{\prime }$ is increasing.

So, ${}^{\prime }{T}^{\prime }$ is also increasing $\Rightarrow$ Internal energy increasing (2)

and $dw=+ve$ ( work is done by the gas )

By first law of thermodynamics,

$\Rightarrow dQ=+ve$ heat gain (4).

(D) Process $D\to A.$

Hence, solved.