JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 24)
One mole of an ideal monoatomic gas is taken along the path ABCA as show in the PV diagram. The maximum temperature attained by the gas along the path BC is given by :
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$${{25} \over {16}}\,{{{P_o}{V_o}} \over R}$$
$${{25} \over {8}}\,{{{P_o}{V_o}} \over R}$$
$${{25} \over {4}}\,{{{P_o}{V_o}} \over R}$$
$${{5} \over {8}}\,{{{P_o}{V_o}} \over R}$$
Penjelasan
Equation of line BC,
P = P0 $$-$$ $${{2{P_0}} \over {{V_0}}}\left( {V - 2{V_0}} \right)$$
As $$\,\,\,\,\,$$ PV = nRT
$$\therefore\,\,\,\,$$ T = $${{PV} \over {nR}}$$
= $${{{P_0}V - {{2{P_0}{V_2}} \over {{V_0}}} + 4{P_0}V} \over {1 \times R}}$$
(As given n = 1 mole gas)
$$ \Rightarrow $$$$\,\,\,$$ T = $${{{P_0}} \over R}\left[ {5V - {{2{V^2}} \over {{V_0}}}} \right]$$
For maximum value of T
$${{dT} \over {dV}} = 0$$
$$\therefore\,\,\,$$ 5 $$-$$ $${{4V} \over {{V_0}}} = 0$$
$$ \Rightarrow $$$$\,\,\,$$ V = $${5 \over 4}{V_0}$$
$$\therefore\,\,\,$$ Tmax = $${{{P_0}} \over R}\left[ {5 \times {{5{V_0}} \over 4} - {2 \over {{V_0}}} \times {{25} \over {16}}{V_0}^2} \right]$$
= $${{25} \over 8} \times {{{P_0}\,{V_0}} \over R}$$
P = P0 $$-$$ $${{2{P_0}} \over {{V_0}}}\left( {V - 2{V_0}} \right)$$
As $$\,\,\,\,\,$$ PV = nRT
$$\therefore\,\,\,\,$$ T = $${{PV} \over {nR}}$$
= $${{{P_0}V - {{2{P_0}{V_2}} \over {{V_0}}} + 4{P_0}V} \over {1 \times R}}$$
(As given n = 1 mole gas)
$$ \Rightarrow $$$$\,\,\,$$ T = $${{{P_0}} \over R}\left[ {5V - {{2{V^2}} \over {{V_0}}}} \right]$$
For maximum value of T
$${{dT} \over {dV}} = 0$$
$$\therefore\,\,\,$$ 5 $$-$$ $${{4V} \over {{V_0}}} = 0$$
$$ \Rightarrow $$$$\,\,\,$$ V = $${5 \over 4}{V_0}$$
$$\therefore\,\,\,$$ Tmax = $${{{P_0}} \over R}\left[ {5 \times {{5{V_0}} \over 4} - {2 \over {{V_0}}} \times {{25} \over {16}}{V_0}^2} \right]$$
= $${{25} \over 8} \times {{{P_0}\,{V_0}} \over R}$$