InterviewSolution
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InterviewSolution
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Air is contained in a piston-cylinder arrangement as shown in Fig. with a cross-sectional area of `4 cm^(2)` and an initial volume of `20 cc`. The air is initially at a pressure of 1 atm and temperature of `20^(@)C`. The piston is connected to a spring whose spring constant is `k = 10^(4) N//m`, and the spring is initially underformed. How much heat must be added to the air inside cylinder to increase the pressure to 3 atm (for air, `CV = 718 J//kg.^(@)C`, molecular of air 28.97)? |
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Answer» When pressure changes from 1 atm to 2 atm, the change in pressure `P = 2 atm` `= 2 xx 1 xx 10^(5) N//m^(2)` The force exerted on the piston `F = PA = 2 xx 10^(5) xx 4 xx 10^(-4)` `= 80 N` The compression of the spring `x = (F)/(k) = (80)/(10^(4)) = 0.008 m` The change in volume of air due to displacement of piston by `x` `Delta V = Ax = 4 xx 10^(-4) xx 0.008` `= 3.2 xx 10^(-6) m^(3)` `:.` Final volume, `V_(2) = V_(1) + Delta V` `= 20 xx 10^(-6) + 3.2 xx 10^(-6)` By equation of state `(P_(1) V_(1))/(T_(1)) = (P_(2) V_(2))/(T_(2))` `T_(2) = (P_(2) V_(2) T_(1))/(P_(1) V_(1)) = ((3)/(1)) xx ((23.3 xx 10^(6)))/((20 xx 10^(-6))) xx (273 + 20)` `= 1020 K` The change in internal energy air `Delta U = mC_(v) Delta T` `= (2.38 xx 10^(-5)) xx 718 xx (1020 - 293)` `= 12.42 J` Work done in compressing the spring by `x` `W = (1)/(2) kx^(2) = (10^(4))/(2) xx (0.008)^(2) = 0.32 J` From the first law of thermodynamics `Q = Delta U + W = 12.42 + 0.32 = 1274 J` |
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