40. Consider the following two-step process. Heat is allowed to flow out of an i

Question

40. Consider the following two-step process. Heat is allowed to
flow out of an ideal gas at constant volume so that its
pressure drops from PA = 3.1 atm to 1.4 atm. Then the gas
expands at constant pressure, from a volume of 6.8 L to VC
= 15.1 L where the temperature reaches its original value.
a) Calculate the total work done by the gas in the process.
b) Calculate the change in internal energy of the gas in the process.
c) Calculate the total heat flow into or out of the gas. (Heat flow into the gas is positive.)

Explanation / Answer

work done by the gas along the path AB , w1 = 0 because volume is constatnt work done by the gas along the path BC = w2 = p ( vc- vb)                                                                             =1.4atm( 15.1- 6.8) * 10-3                                                                             = 1.4 * 1.013 * 105 * 8.3 * 10-3                                                                             = 4.68 * 102 J a) total wor done w = w1 + w2                              =  4.68 * 102 J b) here intial and final    temperature is same so change in internal energy u = 0 c) from the first law of tthermodynamics heat flow in the gas                             Q = u +w                                = 4.68 * 102 J                              =  4.68 * 102 J b) here intial and final    temperature is same so change in internal energy u = 0 c) from the first law of tthermodynamics heat flow in the gas                             Q = u +w                                = 4.68 * 102 J

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

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