One mole of an ideal gas in a closed system, initially at 25°c and 10 bar, is first expanded adiabatically, then heated isochorically to reach a final state of 25°c and 1 bar. assuming these processes are mechanically reversible, compute t and p after the adiabatic expansion, and compute q, w, δu, and δh for each step and for the overall process. take cp = (7/2)r and cv = (5/2)r.

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Δ

Explanation:

Hello,

At the first state, the molar volume is:

The volume in both the second and third state:

Now, as it is about an adiabatic process, one remembers the following relationships:

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

- And the temperature:

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

Then the total heat:

- The work for the first process is:

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

Best regards.

law of conservation of matter