A heat engine uses a diatomic gas that follows the pV cycle in the figure. PART
ID: 2106592 • Letter: A
Question
A heat engine uses a diatomic gas that follows the pV cycle in the figure.
PART A
Determine the pressure, volume, and temperature at point 2.
a. p_2 = 256500 Pa, V_2 = 1000 cm^3, T_2 = 845.2 K
b. p_2 = 689600 Pa, V_2 = 1000 cm^3, T_2 = 382.3 K
c. p_2 = 781200 Pa, V_2 = 1000 cm^3, T_2 = 382.3 K
d. p_2 = 696400 Pa, V_2 = 1000 cm^3, T_2 = 522.3 K
PART B
Determine delta E_th, W_s, and Q for the process 1-->2.
a. deltaE_12 = 741.1 J, (W_s)_12 = 0 J, Q_12 = 741.1 J
b. deltaE_12 = 841.1 J, (W_s)_12 = 0 J, Q_12 = 841.1 J
c. deltaE_12 = -741.1 J, (W_s)_12 = 0 J, Q_12 = -741.1 J
d. deltaE_12 = 0 J, (W_s)_12 = 0 J, Q_12 = 741.1 J
Explanation / Answer
for first question volume is constant so:
P1/T1 = P2/T2
400/300 = P2/T2
option d satisfies the relation above
now for question 2
as volume is constant work done will be zero
Q = U
Cv for diatomic gas is (5/2)R
U = (5/2)R*(522.3 - 300)
Q = U
from here get answer for part B
now for part C
process is adiabatic so Q=0
W = -U
W = (P3V3 - P2V2)/(gamma - 1)
gamma for diatomic gas is 7/5
put values and solve
now for part D
process 3 to 1 is isotherm so U=0
W = RT(lnV3/V2)
net work done = W(2-3) + W(3-1)
add both work according to sign which have already been calculated in part above
Efficiency = 1 - (work done)/(heat given)