A gas is taken through the cyclic process described in the figure. (The x axis i

Question

A gas is taken through the cyclic process described in the figure. (The x axis is. marked in increments of 5 m^3.) Find the net energy transferred to the system by heat during one complete cycle. kJ If the cycle is reversed-that is, the process follows the path ACBA-what is the net energy input per cycle by heat? kJ An ideal gas initially .it 335 K undergoes an isobaric expansion at 2.50 kPa. The volume increases from 1.00 m^3 to 3.00 m^3 and 14.4 kJ is transferred to the gas by heat. What is the charge in internal energy of the gas? kJ What is the final temperature of the gas? K

Explanation / Answer

problem 5:

In a cyclic process,heat exchange=Q=W=area enclosed in PV diagram=area triangle ABC=(1/2)6000*4=12000 J

As cycle is clockwise C to A to B ,the work is positive
Heat added to gas is 12000 J
When cycle is reversed , C to B to A cycle traced anticlockwise, heat is negative
Heat =work = -12000 J

Heat extracted from the gas is 12000 J

problem 6:

(a)
Change of Internal energy equals heat transferred to plus work done on the gas
U = Q + W

Work done on the gas by changing Volume from V to V is given by the integral
W = - p dV from V to V
for isobaric process ( p=constant)
W = - p·(V - V)
for this problem
W = - 2.5×10^3Pa · (3m³ - 1m³) = -5kJ

Hence
U = 14.4kJ - 5kJ = 9.4kJ

(b)
Use ideal gas law
p·V = N·R·T
Since pressure p and N stay constant throughout the process:
V·T = p / (N·R) = constant
hence:
V·T = V·T
<=>
T = T· (V/V)
= 335K · (1m³ / 3m³)
= 111.67K

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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