U=120kJ=n∗INTCvdT=n∗INT(Cp−R)dT=n∗INTCpdT−n∗R∗(T2−T1) . Here, R=8.314J/mol−K , n = Mair / MWair and MWair ~ 29g/mol. Cp is a polynomial that you can look up in your book or on the NIST website: Cp=A+(B/1000)∗T+(C/10002)∗T2+(D/10003)∗T3+(E/1000−2)/T2 , where T is in Kelvin and Cp is in J/mol-K. The values of the constants are: A=26.84107,B=7.7816776,C=−1.8103208,D=0.14594026,E=−0.01102637. You can integrate this polynomial fairly easily.
But we do not know T1 or T2. Use the ideal gas equation of state to determine T1:P1V1=nRT1.So,T1=P1∗V1/(n∗R).IgotT1=348.81K=75.66C.
The problem is that T2 is the ONLY unknown in the equation. You could use SOLVER in EXCEL or a more powerful math package to determine a value of T2 that yields a ?U of 120 kJ. Once you know T2, use the ideal gas equation of state to determine V2:P2V2=nRT2orV2=nRT2/P2.
An approximate solution could be obtained if you assume the Cp of air is a constant, say about 1 kJ/kg-K. In this case, ?U = m Cp * (T2-T1). Solving for T2 yields: T2 = T1 + ?U/(m Cp). Plugging in values yields: T2 = 468.81 K = 195.66C.
Finally, V2 = 0.02688 m^3.
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