Saved Bookmarks
| 1. |
Find the efficiency of the thermo-dynamic cycle shown in figure for an ideal diatomic gas. |
|
Answer» Solution :Let n be the number of moles of the gas and the temperature be `T_0` in the state A. Now, work done during the cycle `= W=1/2 xx (2 V_0-V_0) ( 2P_0 -P_0) =1/2P_0 V_0` For the heat `(Delta Q_1)` GIVEN during the process `A toB`, we have ` Delta Q_1- Delta W_(AB ) +Delta U_(AB )` `Delta W_(AB ) =` area under the straight line `AB =1/2(P_0 +2P_0 ) (2V_0-V_0) =(3 P_0 V_0)/(2)` Applying EQUATION of state for the gas in the state A and B. `(P_0V_0) /(T_0 ) = ((2P_0) (2V_0))/(T_B) =T_B= 4T_0thereforeU_(AB )= nC_v =Delta T =n ((5R)/( 2) ) (4 T_0-T_0 ) = ( 15nRT_0)/( 2) = ( 15P_0 V_0)/(2)` ` thereforeDeltaQ_1= 3/2P_0V_0+ (15)/(2)P_0 V_0= 9 P_0 V_0` Obviously, the processes B `to` C and C `to` A involve the extraction of heat from the gas. ` therefore ` Efficiency (`eta `) of the cycle `=("WORKDONE percycle ")/( "Total heat supplied per cycle")i.e., eta =(1/2 P_0 V_0 )/(9 p_0 V_0) = 1/(18)` |
|