InterviewSolution
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InterviewSolution
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An adiabatic vessel of total volume V is divided into two equal parts by a conducting separtor. The is fixed in this position. The part on the left contains one mole of an ideal gas `(U=1.5 nRT)` and the part on the right contains two moles of the same gas. initially, the pressure on each side is p. the system is left for sufficiant time so that a steady state is reached. find (a) the work done by the gas in the left part during the process, (b) the temperature on the two sides in the beginning, (c ) the final common temperature reached by the gases, (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part. |
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Answer» Correct Answer - A::B::C::D (a) As the conducting wall is fixed the work done by the gas on the left part during the process is zero. (b) For left side, Let pressure `= P` `Volume = V` No of moles `= n (1 mole)` Let initial temperature `= T_(1) `(PV)/(2) = nRT_1` `rArr (PV)/(2) = (1)nRT` `rArr T_1 = (PV)/((2moles) R)` For Rightside No of moles` = (2moles). ` ` Let initial temperature` = T_2` (PV)/(2) = nRT_(2)` `T_2 = (PV)/(2moles)R)` Let final temperature = t Final pressure `= P` No of moles ` = 1mole+2mole` ` =3mole` `PV =nRT` ` ` ( PV)/(nR)` ` ` = (PV)/((3mole)xxR)` (d) For RHS, ` ` Delta Q =Delta U as Delta W = 0` ` Delta U = 1.5n_2R(T-T_2)` ` = 1.5xx2xxR(T-T_2)` ` 1.5xx2xx(4PV-3PV)/(4xx3 mole)` ` = (3xxPV)/(4xx3 moles) = (PV)/(4)` (e) As `dQ = -dU` ` rArr dU = -dQ = (-PV)/(4)` |
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