- #1
Allison Barry
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Homework Statement
A specific type of ideal gas has a specific heat capacity at constant pressure (cp=cv+R) that is a function of temperature T, such that cp=0.5+876T, where cp has units of J/kg/K and T has units of K. The gas, which is initially at T1 = 294 K and P1 = 1x105 Pa, undergoes a reversible adiabatic process such that its final temperature is T2 = 778 K. Calculate the pressure of the gas (in Pa) in this final state. Assume the following ideal gas constant: R = 287 J/kg/K. Recall that ds = cpdT/T – RdP/P.
Relevant equations
Reversible Adiabatic process:
Entropy is 0 because this is adiabatic and reversible.
Constituitive relation for entropy of such a process is:
S2 - S1 = ∫12dS = 0
dS = cp(dT/T) - R(dP/P) = 0
Attempt:
Integrating I got:
∫T1T2(0.5 + 876T).(dT/T) - 287∫P1P2(dP/P)
876(778 - 294) + (0.5) ln (778/294) = 287ln(P2 / 100,000Pa)
Getting 1488.810788 = lnP2
But e1488.810788 is undefinable? I am confused? Did I integrate this wrong?
A specific type of ideal gas has a specific heat capacity at constant pressure (cp=cv+R) that is a function of temperature T, such that cp=0.5+876T, where cp has units of J/kg/K and T has units of K. The gas, which is initially at T1 = 294 K and P1 = 1x105 Pa, undergoes a reversible adiabatic process such that its final temperature is T2 = 778 K. Calculate the pressure of the gas (in Pa) in this final state. Assume the following ideal gas constant: R = 287 J/kg/K. Recall that ds = cpdT/T – RdP/P.
Relevant equations
Reversible Adiabatic process:
Entropy is 0 because this is adiabatic and reversible.
Constituitive relation for entropy of such a process is:
S2 - S1 = ∫12dS = 0
dS = cp(dT/T) - R(dP/P) = 0
Attempt:
Integrating I got:
∫T1T2(0.5 + 876T).(dT/T) - 287∫P1P2(dP/P)
876(778 - 294) + (0.5) ln (778/294) = 287ln(P2 / 100,000Pa)
Getting 1488.810788 = lnP2
But e1488.810788 is undefinable? I am confused? Did I integrate this wrong?