Thermodynamics-Finding Internal energy

[tre2k3]

I am going over this example

0.5 kg of air undergoes a Carnot cycle with η = 0.5.
Given the initial pressure p1 = 700 kPa, initial
volume V1 = 0.12 m3 and heat transfer during the
isothermal expansion process Q12 = 40 kJ, Find
the highest and the lowest temperatures in the
cycle.
(b) the amount of heat rejection.
(c) work in each process.

They have Th to be 585.4k and Tl to be 292.7k
and so
W23 = m(u2 – u3)= 0.5(423.7 – 208.8) = 107.5 kJ

I do not know how to findu2 and u3. Looking at the tables at he back of the book doesn't help because they don't list the temperatures in this problem. So what equation is used to find u2 and u3 with the given information.

 

[physixguru]

What is the work done during an isothermal process?

What is the efficiency of a carnot cycle?

 

[Moetasim]

I would first like to clarify that the given information is not enough to accurately determine the internal energy at states 2 and 3 (u2 and u3). The internal energy of a gas depends on its temperature, pressure, and specific heat capacity, and in this case, we only have the initial pressure and volume, and the highest and lowest temperatures in the cycle.

To accurately determine the internal energy at states 2 and 3, we would need to know the specific heat capacity of the air at those temperatures. This information is not provided in the given example, so it is not possible to calculate u2 and u3.

However, in general, the internal energy of a gas can be calculated using the following equation:

u = cvT + u0

Where:
u = internal energy
cv = specific heat capacity at constant volume
T = temperature
u0 = constant specific internal energy at a reference state

In this case, we would need to know the specific heat capacity at constant volume for air at the given temperatures to calculate u2 and u3.

Regarding the other questions, the highest and lowest temperatures can be determined using the Carnot efficiency equation:

η = (Th - Tl)/Th

Where:
η = Carnot efficiency
Th = highest temperature
Tl = lowest temperature

To find the amount of heat rejection (Q23), we can use the first law of thermodynamics:

Q23 = W23 + ΔU23

Where:
Q23 = heat rejected
W23 = work done in the process
ΔU23 = change in internal energy

We can calculate the work done in each process using the following equations:

W12 = Q12 = 40 kJ (isothermal process)
W23 = m(u2 - u3) (adiabatic process)
W34 = Q34 + W23 = Q34 + m(u3 - u4) (isothermal process)

I hope this explanation helps in understanding how to approach this problem. However, without the specific heat capacity of air at the given temperatures, it is not possible to accurately determine the internal energy at states 2 and 3.