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

Question

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 = 3.7 atm to 1.4 atm. Then the gas expands at constant pressure, from a volume of 6.8 L to VC = 18.0 L where the temperature reaches its original value. See the figure below.

(a) Calculate the total work done by the gas in the process.
_________ J
(b) Calculate the change in internal energy of the gas in the process.
_________J
(c) Calculate the total heat flow into or out of the gas. (Heat flow into the gas is positive.)
_________J

Explanation / Answer

a) no work form a to b

wtot = wbc = p(v2-v1) = 1.4*1.01*10^5*(18-6.8)*10^-3 = 1583.7 J

b) Since the temperature at the beginning and at the end of the process are the same, there is no change in internal energy

c) from first law of thermodynamics Q = u+w = w = 1583.7 J into the gas

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects