A heat engine takes 0.350 mol of an ideal diatomic gas around the cycle shown in
ID: 1568336 • Letter: A
Question
A heat engine takes 0.350 mol of an ideal diatomic gas around the cycle shown in the pV diagram of (Figure 1) . Process 12 is at constant volume, process 23 is adiabatic, and process 31 is at a constant pressure of 1.00 atm. The value of for this gas is 1.40. The magnitude of the change in internal energy for each process is |U12|=2180J, |U23|=785J, and |U31|=1396J.
C. Calculate Q for each of the three processes. (Express your answers in joules to three significant figures, separated by commas)
D. Calculate W for each of the three processes. (Express your answers in joules to three significant figures, separated by commas)
1.00 atm 1 TI 300 K T 492 KExplanation / Answer
Given that
number of moles n=0.350
degrees of freedom =diatomic
gamma=7/5
now we find the change in heat energy in each process
process1=>the heat energy dQ1=nCvdT=0.350*5/2*8.3[600-300]=2178.8 J
process 2=>the heat energy dQ2=0 J because in the adiabatic process there is no exchange of heat energy
process 3=>the heat energy dQ3=nCpdt=0.350*7/2*8.3*[-492+300]=-1952.2 J
now we find the amount of work done in each process
process1=>work done W1=0 because the volume is constant
process 2=>work done W2=nR[T1-T2]/1.4-1=0.350*8.3[600-492]/0.4=784.4 J
process 3=>the work done W3=nRdT=0.350*8.3*[-492+300]=-557.8 J