Question 343: Thermodynamics & KTG

Question

If one mole of an ideal gas at (P
1
, V
1
) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B $ 	o $ C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net workdone by the gas is equal to :
![](https://app-content.cdn.examgoal.net/fly/@width/image/1la259wx9/64c188df-6d15-42cc-b5ee-f5e4aa79ad51/fa6819d0-5c0d-11ed-b429-1bdcfc436fa3/file-1la259wxa.png?format=png)

Accepted Answer

Let p
i
, p
f
, V
i
and V
f
be the initial and final pressure and volume.
Given, AB is isothermal ($\Delta$T = 0),
BC is isochoric ($\Delta$V = 0) and CA is adiabatic ($\Delta$Q = 0)
Since, isothermal work (W
AB
) = ${p_1}{V_1}\ln {{{V_f}} \over {{V_i}}}$
where, V
i
and V
f
are volume at A and B, respectively.
$	herefore$ ${W_{AB}} = {p_1}{V_1}\ln {{2{V_1}} \over {{V_1}}} = {p_1}{V_1}\ln 2$
Since, at constant volume, work done is zero.
$	herefore$ ${W_{BC}} = 0$
Since, W
CA
is an adiabatic work done, i.e.
${W_{CA}} = {1 \over {1 - \gamma }}({p_f}{V_f} - {p_i}{V_i})$
$ \Rightarrow {W_{CA}} = {1 \over {1 - \gamma }}({p_1}{V_1} - {{{p_1}} \over 4} 	imes 2{V_1})$
$ = {1 \over {1 - \gamma }}({p_1}{V_1} - {p_1}{V_1}/2) = {1 \over {1 - \gamma }}{{{p_1}{V_1}} \over 2}$
$	herefore$ Net work done, ${W_{net}} = {W_{AB}} + {W_{BC}} + {W_{CA}}$
$ = {p_1}{V_1}\ln 2 + 0 + {1 \over {1 - \gamma }}{{{p_1}{V_1}} \over 2}$
$ = {p_1}{V_1}[\ln 2 + 1/2(1 - \gamma )]$
From ideal gas law, $pV = nRT$
$	herefore$ ${W_{net}} = RT[\ln 2 - 1/2(\gamma  - 1)]$ ($\because$ n = 1)
![](https://app-content.cdn.examgoal.net/fly/@width/image/1kxkkm69d/6cdc1166-b9cb-470d-811e-514eb62d9bb1/44078a10-64d1-11ec-b334-61c7774f09f5/file-1kxkkm69e.png?format=png)

Answer

$ - {{RT} \over {2(\gamma  - 1)}}$

Answer

$RT\left( {\ln 2 - {1 \over {2(\gamma  - 1)}}} \right)$

Answer

$RT\ln 2$

Answer

$0$

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