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.

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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explanation:

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge