A heat engine takes 0.310 mol of an ideal diatomic gas around the cycle shown in
ID: 2218706 • Letter: A
Question
A heat engine takes 0.310 mol of an ideal diatomic gas around the cycle shown in the pV diagram of the figure. Process 1,2 : Constant volume, process 2,3: adiabatic. Process 3,1: Constant Pressure (1 atm). Value of Y for this gas is 1.40
Calculate Q for each of 3 processes: Q(1,2) , Q(2,3) , Q(3,1)
For Q(1,2), the answer is somewhat close to 1878 J. However, when I put in 1933 J, it doesn't say its close.
A heat engine takes 0.310 mol of an ideal diatomic gas around the cycle shown in the pV diagram of the figure. Process 1,2 : Constant volume, process 2,3: adiabatic. Process 3,1: Constant Pressure (1 atm). Value of Y for this gas is 1.40 Calculate Q for each of 3 processes: Q(1,2) , Q(2,3) , Q(3,1) For Q(1,2), the answer is somewhat close to 1878 J. However, when I put in 1933 J, it doesn't say its close.Explanation / Answer
the cycle has 3 different processes
using ideal gas law V=nRT/P
at pos 1.V=900R/P=0.0738m^3
at pos 2.V=0.1211m^3
at pos 3.V=0.0738m^3
a.isobaric expansion
work done=PV=101325*0.0473=-4790J..approx
ii.isochoric transition from pos1->2
work done=0
iii.adiabatic change from pos 2->3
work done=nCv(TfTi)
Cv for diatomic gas=2.5R
therefore W=6734J
total work6734-4790=1944J.......ANS