Why doesn't (P1xV1)/T1 = (P2xV2)/T2 work here?

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The discussion centers on the application of the ideal gas law, specifically the equation (P1 x V1)/T1 = (P2 x V2)/T2, in a scenario involving R-134a in a rigid container. The initial conditions include 10 kg of R-134a at -40 degrees Celsius, with the container heated to a pressure of 200 kPa. The calculated final temperature is 66.3 degrees Celsius, and the initial pressure is 51.25 kPa, derived from thermodynamic property tables. The equation fails because R-134a does not behave as an ideal gas under these conditions, particularly due to phase changes and non-ideal behavior at varying temperatures and pressures. Understanding the limitations of the ideal gas law is crucial when dealing with real gases like R-134a in thermodynamic processes.
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10 kg of R-134a fill a 1.348 m^3 rigid container at an initial temperature of -40 degrees Celsius. The container is then heated until the pressure is 200 kPa. Determine the final temperature and the initial pressure.

I found the final temperature to be 66.3 degrees Celsius and the initial pressure to be 51.25 kPa using thermodynamic property tables. However, when I plugged all of the values into the equation (P1 x V1)/T1 = (P2xV2)/T2, it does not work. Can anyone explain why?
 
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Would you expect it to behave as an ideal gas over that temperature range?
 
ahh. i see. thanks =)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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