JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 7)
If one mole of an ideal gas at (P1, V1) 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 $$ \to $$ 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 :
_24th_February_Evening_Shift_en_7_1.png)
$$ - {{RT} \over {2(\gamma - 1)}}$$
$$RT\left( {\ln 2 - {1 \over {2(\gamma - 1)}}} \right)$$
$$RT\ln 2$$
$$0$$
Explanation
Let pi, pf, Vi and Vf 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 (WAB) = $${p_1}{V_1}\ln {{{V_f}} \over {{V_i}}}$$
_24th_February_Evening_Shift_en_7_2.png)
where, Vi and Vf are volume at A and B, respectively.
$$\therefore$$ $${W_{AB}} = {p_1}{V_1}\ln {{2{V_1}} \over {{V_1}}} = {p_1}{V_1}\ln 2$$
Since, at constant volume, work done is zero.
$$\therefore$$ $${W_{BC}} = 0$$
Since, WCA is an adiabatic work done, i.e.
$${W_{CA}} = {1 \over {1 - \gamma }}({p_f}{V_f} - {p_i}{V_i})$$
$$ \Rightarrow {W_{CA}} = {1 \over {1 - \gamma }}({p_1}{V_1} - {{{p_1}} \over 4} \times 2{V_1})$$
$$ = {1 \over {1 - \gamma }}({p_1}{V_1} - {p_1}{V_1}/2) = {1 \over {1 - \gamma }}{{{p_1}{V_1}} \over 2}$$
$$\therefore$$ 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, $$pV = nRT$$
$$\therefore$$ $${W_{net}} = RT[\ln 2 - 1/2(\gamma - 1)]$$ ($$\because$$ n = 1)
Given, AB is isothermal ($$\Delta$$T = 0),
BC is isochoric ($$\Delta$$V = 0) and CA is adiabatic ($$\Delta$$Q = 0)
Since, isothermal work (WAB) = $${p_1}{V_1}\ln {{{V_f}} \over {{V_i}}}$$
where, Vi and Vf are volume at A and B, respectively.
$$\therefore$$ $${W_{AB}} = {p_1}{V_1}\ln {{2{V_1}} \over {{V_1}}} = {p_1}{V_1}\ln 2$$
Since, at constant volume, work done is zero.
$$\therefore$$ $${W_{BC}} = 0$$
Since, WCA is an adiabatic work done, i.e.
$${W_{CA}} = {1 \over {1 - \gamma }}({p_f}{V_f} - {p_i}{V_i})$$
$$ \Rightarrow {W_{CA}} = {1 \over {1 - \gamma }}({p_1}{V_1} - {{{p_1}} \over 4} \times 2{V_1})$$
$$ = {1 \over {1 - \gamma }}({p_1}{V_1} - {p_1}{V_1}/2) = {1 \over {1 - \gamma }}{{{p_1}{V_1}} \over 2}$$
$$\therefore$$ 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, $$pV = nRT$$
$$\therefore$$ $${W_{net}} = RT[\ln 2 - 1/2(\gamma - 1)]$$ ($$\because$$ n = 1)
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