JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 26)
A sample of gas with $$\gamma$$ = 1.5 is taken through an adiabatic process in which the volume is compressed from 1200 cm3 to 300 cm3. If the initial pressure is 200 kPa. The absolute value of the workdone by the gas in the process = _____________ J.
Answer
480
Explanation
v = 1.5
p1v1v = p2v2v
(200) (1200)1.5 = P2 (300)1.5
P2 = 200 [4]3/2 = 1600 kPa
$$\left| {W.D.} \right| = {{{p_2}{v_2} - {p_1}{v_1}} \over {v - 1}} = \left( {{{480 - 240} \over {0.5}}} \right) = 480$$ J
p1v1v = p2v2v
(200) (1200)1.5 = P2 (300)1.5
P2 = 200 [4]3/2 = 1600 kPa
$$\left| {W.D.} \right| = {{{p_2}{v_2} - {p_1}{v_1}} \over {v - 1}} = \left( {{{480 - 240} \over {0.5}}} \right) = 480$$ J
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