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

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)

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.