One mole of an ideal monatomic gas undergoes a cycle as shown in the figure. Ass

Question

One mole of an ideal monatomic gas undergoes a cycle as shown in the figure. Assume the system to be defined as the gas in the following problem. The temperatures at the 3 points are TA=325.0 K, TB=740.0 K, and Tc=532.5 K. Process AB oocurs at constant volume, process BC is adiabatic, and process C-A occurs at constant pressure. For process AB, what is the heat, Q, absorbed by the system? Tries 1/10Previous Tries Submit Answer For process AB, what is the change in internal energy, Submit Answer Tries 0/10 For process AB, what is the work, W, done by the system? You are correct Your receipt no. is 158-238 For process BC, what is the heat, Q, absorbed by the system? 0 J You are correct. Your receipt no. is 158-2054 For process BC, what is the change in internal energy, Eint? Submit Answer Tries 0/10 For process B C, what is the work, W, done by the system? Submit Answer Tries 0/10 For process CA, what is the heat, Q, absorbed by the system? Submit Answer Incorrect. Tries 1/10Previous Tries For process CA, what is the change in internal energy, Eint? Submit Answer Incorrect. Tries 1/10previous Tries For process CA, what is the work, W, done by the system? Submit Answer Tries 0/10 what is Eint, the change in internal energy, for the full cycle ABCA? Submit Answer Trles 0/10 What is W, the work done by the system, for the full cycle ABCA? Submit Answer Tries 0/10 What is Q, the heat absorbed by the system, for the full cycle ABCA? Submit Answer Tries 0/10 For the following questions assume the initial pressure at Point A is (-1.013×105 Pa) and assume R-8.31 What is the volume at Point B? Submit Answer Tries o/10 What t is the pressure at Point B? Submit Answer Tries 0/10 What is the volume at Point C?

Explanation / Answer

A)

for A--> B

this is an isochoric process, since TA < Tb, then there is absorbtion of heat

Q = n*C+(Tf-Ti)

Q = 1*(3/2R)*(740-325)

Q = 3/2*8.314*(740-325) = 5175.465 J

b)

find change in internal energy

since W = 0, then

Q = dU

then

dU = 5175.465 J

C

for A--> B, dV = 0, so W = -P*dV = 0

D)

for B-->C, there is no heat, since this is isothermal

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