JEE MAIN - Physics (2013 (Offline) - No. 21)
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The above $$p$$-$$v$$ diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is
$${p_0}{v_0}$$
$$\left( {{{13} \over 2}} \right){p_0}{v_0}$$
$$\left( {{{11} \over 2}} \right){p_0}{v_0}$$
$$4{p_0}{v_0}$$
Explanation
Along path DA, volume is constant.
Hence, $$\Delta $$QDA = nCv$$\Delta $$T = nCv(TA – TD)
$$ \therefore $$ $$\Delta $$QDA = $$n\left( {{3 \over 2}R} \right)\left[ {{{2{p_0}{v_0}} \over {nR}} - {{{p_0}{v_0}} \over {nR}}} \right] = {3 \over 2}{p_0}{v_0}$$
Along the path AB, pressure is constant.
Hence $$\Delta $$QAB = nCp$$\Delta $$T = nCp(TB – TA)
$$ \therefore $$ $$\Delta $$QAB = $$n\left( {{5 \over 2}R} \right)\left[ {{{2{p_0}2{v_0}} \over {nR}} - {{2{p_0}{v_0}} \over {nR}}} \right] = {{10} \over 2}{p_0}{v_0}$$
$$ \therefore $$ The amount of heat extracted from the source in a single cycle is
$$\Delta $$Q = $$\Delta $$QDA + $$\Delta $$QAB
$$ = {3 \over 2}{p_0}{v_0} + {{10} \over 2}{p_0}{v_0}$$ = $$\left( {{{13} \over 2}} \right){p_0}{v_0}$$
Hence, $$\Delta $$QDA = nCv$$\Delta $$T = nCv(TA – TD)
$$ \therefore $$ $$\Delta $$QDA = $$n\left( {{3 \over 2}R} \right)\left[ {{{2{p_0}{v_0}} \over {nR}} - {{{p_0}{v_0}} \over {nR}}} \right] = {3 \over 2}{p_0}{v_0}$$
Along the path AB, pressure is constant.
Hence $$\Delta $$QAB = nCp$$\Delta $$T = nCp(TB – TA)
$$ \therefore $$ $$\Delta $$QAB = $$n\left( {{5 \over 2}R} \right)\left[ {{{2{p_0}2{v_0}} \over {nR}} - {{2{p_0}{v_0}} \over {nR}}} \right] = {{10} \over 2}{p_0}{v_0}$$
$$ \therefore $$ The amount of heat extracted from the source in a single cycle is
$$\Delta $$Q = $$\Delta $$QDA + $$\Delta $$QAB
$$ = {3 \over 2}{p_0}{v_0} + {{10} \over 2}{p_0}{v_0}$$ = $$\left( {{{13} \over 2}} \right){p_0}{v_0}$$
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