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.

Δ

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.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

elements make up everything, and they combine to make compounds.

the answer is c