JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 11)
Explanation
For monoatomic gas degree of freedom f = 3 and $$\gamma$$ = $${5 \over 3}$$
Here for gas,
Initial pressure (P1) = 75 kPa
Initial volume (V1) = 1200 cm3
Final volume (V2) = 150 cm3
Final pressure (P2) = ?
For adiabatic process,
$${P_1}V_1^\gamma = {P_2}V_2^\gamma $$
$$ \Rightarrow 75 \times {(1200)^\gamma } = {P_2}{(150)^\gamma }$$
$$ \Rightarrow {P_2} = 75 \times {\left( {{{1200} \over {150}}} \right)^{{5 \over 3}}}$$
$$ = 75 \times {(8)^{{5 \over 3}}}$$
$$ = 75 \times 32$$ kPa
= 2400 kPa
Work done in adiabatic process,
$$W = {{{P_2}{V_2} - {P_1}{V_1}} \over {1 - \gamma }}$$
$$ = {{2400 \times {{10}^3} \times 150 \times {{10}^{ - 6}} - 75 \times {{10}^3} \times 1200 \times {{10}^{ - 6}}} \over {1 - {5 \over 3}}}$$
$$ = {{(2400 \times 150 - 75 \times 1200) \times {{10}^{ - 3}}} \over { - {2 \over 3}}}$$
$$ = {{ - 810000 \times {{10}^{ - 3}}} \over 2}$$
$$ = - 405$$ kJ
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