Show the dissolution of KCl, HBr and Methanol via structures also name the type of interaction which helps in its dissolution in
1 answer:
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Explanation:
- For path A, the calculation will be as follows.
As, for reversible isothermal expansion the formula is as follows.
W =
Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.
As the given data is as follows.
R = 8.314 J/(K mol), T = 298 K ,
= 8.34 L,
= 2.67 L
Now, putting the given values into the above formula as follows.
W =
=
=
= -2818.68 J
Hence, work for path A is -2818.68 J.
- For path B, the calculation will be as follows.
Step 1: When there is no change in volume then W = 0
Hence, for step 1, W = 0
Step 2: As, the gas is allowed to expand against constant external pressure = 1.00 atm.
So, W =
Now, putting the given values into the above formula as follows.
W =
=
= -5.67 atm L
As we known that, 1 atm L = 101.33 J
Hence, work will be calculated as follows.
W =
= -574.54 J
Therefore, total work done by path B = 0 + (-574.54 J)
W = -574.54 J
Hence, work for path B is -574.54 J.