Thermal Machine: Solving for Q23 to Complete the Otto Cycle

AI Thread Summary
The discussion focuses on solving for Q23 in the Otto cycle, specifically addressing the transitions between states 1 to 2 (adiabatic), 2 to 3 (isochoric), 3 to 4 (adiabatic), and 4 to 1 (isochoric). The equations provided relate pressure, volume, and temperature across these states, with specific values for P2 and T2 given. The challenge arises in determining the values for P3, P4, T3, and T4, as the equations appear interdependent. The conclusion suggests that rather than missing data, the values should be expressed in terms of Q23 to complete the cycle. The goal is to find the value of Q23 that allows the cycle to close at point 1.
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Homework Statement
Consider an Otto Cycle:
1-->2 (adiabatic)
2-->3 (isochoric)
3-->4 (adiabatic)
4-->1 (isochoric)

Knowing that:

P1 = 10^5 Pa
V1= 0,8 m^3
T1 = 290K
V3= 0,1m^3
cv= 0,171 Kcal/kg . K
Gas mass = 1 kg
gamma (coefficient of adiabatic expansion) = 1,4
V1=V4
V2=V3
Q23 = heat received
Q41 = heat expelled

Find T2,T3,T4 and P2,P3,P4
Relevant Equations
pv/t = constant
p(v)^gamma= constant
t v^(1-gamma) = constant
Q = m. cv . (T'-T)
1-->2 (adiabatic)

$$P_1V_1^γ=P_2V_2^γ$$

$$P_2=18,4.10^5Pa$$

$$T_1V_1^{1-\gamma}=T_2V_2^{1-\gamma}$$

$$T_2=429,32K$$

2-->3 (isochoric)

$$\frac{P_2}{T_2}=\frac{P_3}{T_3}$$

$$\frac{18,4.10^5}{429,32}=\frac{P_3}{T_3}$$

3--> 4 (adiabatic)

$$P_3V_3^{\gamma}=P_4V_4^{\gamma}$$

$$T_3V_3^{\gamma-1}=T_4V_4^{\gamma-1}$$

4--> 1 (isochoric)

$$\frac{P_1}{T_1}=\frac{P_4}{T_4}$$

From here, I can no longer find the values of P3, P4, T3, T4, as it is as if each equation is an association of the other. So I was wondering if there is a way to resolve this issue, because for me, I believe that data is missing.
 
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Data is not missing. Express everything for points 3 and 4 in terms of Q23. Solve for the value of Q23 that closes the loop at point 1.
 
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