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26 April, 05:01

A mass of 1 kilogram of air in a piston-cylinder assembly undergoes two processes in series from an initial state where p1 = 1.1 MPa, T1 = 299°C:

Process 1-2: Constant-temperature expansion until the volume is twice the initial volume.

Process 2-3: Constant-volume heating until the pressure is again 1.1 MPa. Assuming ideal gas behavior, determine the temperature at state 3, in kelvin.

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  1. 26 April, 05:42
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    T3 = 1144 K

    Explanation:

    In process 1, we are given;

    P1 = 1.1 MPa = 1100 KPa

    T1 = 299°C and it's constant

    We are told that Final volume is twice the initial volume.

    Thus, V2 = 2V1

    Now, from ideal gas equation, using Boyles law we know that;

    P1•V1 = P2•V2

    Thus, let's make P2 the subject;

    P2 = P1•V1/V2

    V2 = 2V1

    Thus;

    P2 = P1•V1 / (2V1)

    V1 will cancel out to give;

    P2 = P1/2

    Thus, P2 = 1100/2

    P2 = 550 KPa

    Now, for process 2-3;

    The temperature in process 1 was constant and thus T1 = T2 = 299°C. Also, since ideal gas behavior, P3 = P1 = 1100 KPa

    Since, we are to find the temperature T3 in kelvins.

    Let's convert T2 into kelvins.

    So, T2 = 299 + 273 K = 572 K

    So, from ideal gas equations again;

    But this time volume is the one that is constant. Thus;

    P2/T2 = P3/T3

    T3 = P3•T2/P2

    Thus;

    T3 = 1100•572/550

    T3 = 1144 K
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