InterviewSolution
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InterviewSolution
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A cylindrical container having non-conducting walls is partitioned in two equal parts such that the volume of the each parts is eqaul to `V_(0)` A movable non-conducting piston is kept between the two parts. Gas on left is slowly heated so that the gas on right is compressed upto volume `V_(0)//8`. Find pressure and temperature on both sides if initial pressure and temperature, were `P_(0)` and `T_(0)` respectively. Also find heat given by the heater to the gas (number of moles in each parts is n) |
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Answer» Since the process on the right is adiabatic therefore `PV^(gamma)` = constant `implies P_(0) V_(0)^(gamma) = P_("final ") ((V_(0))/(8))^(gamma) implies P_("final") = 32 P_(0)` `T_(0) V_(0)^(gamma - 1) = T_("final") ((V_(0))/(8))^(gamma -1) implies T_("final" = 4 T_(0)` Let volume of the left part is `V_(1)` `implies 2V_(0) = V_(1) + (V_(0))/(8) implies V_(1) = (15 V_(0))/(8)` Since number of moles on the left parts remains constant therefore the left part `(PV)/(T)` = constant. Final pressure on both sides will be same `implies (P_(0) V_(0))/(T_(0)) = (P_("final") V_(1))/(T_("final")) implies T_("final") = 60 T_(0)` `Delta Q = Delta u + Delta W` `Delta Q = (5 R)/(2) (60 t_(0) - T_(0)) + n (3R)/(2) (4T_(0) - T_(0))` `Delta Q = (5 nR)/(2) xx 59 T_(0) + (3 nR)/(2) xx 3T_(0)` |
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