For liquid water at 25°C, α = 2.07 x 10^-4 K^-1; the density may be taken as 1.00 g/cm^3.

For liquid water at 25°C, α = 2.07 x 10^-4 K^-1; the density may be taken as 1.00 g/cm^3. One mole of liquid water is compressed isothermally, 25°C, from 1 atm to 1000 atm. Calculate ΔS, a) supposing that water is incompressible; that is, K = 0. b) supposing that K = 4.53 x 10^-5 atm^-1. answers a) -0.377 J/K mol b) -0.369 J/K mol

The Correct Answer and Explanation is:

To solve this problem, we need to calculate the change in entropy (ΔS\Delta SΔS) for the compression of one mole of water isothermally at 25°C. There are two cases to consider: one where water is incompressible and one where the compressibility factor KKK is given.

General Formulation for ΔS\Delta SΔS

The change in entropy for an isothermal compression process can be determined by the formula:ΔS=nRln⁡(VfVi)\Delta S = nR \ln\left(\frac{V_f}{V_i}\right)ΔS=nRln(Vi​Vf​​)

where:

• nnn is the number of moles (1 mole in this case),
• RRR is the universal gas constant (8.314 J/mol·K),
• VfV_fVf​ and ViV_iVi​ are the final and initial volumes, respectively.

Now, we can calculate the volumes using the ideal gas law:V=nRTPV = \frac{nRT}{P}V=PnRT​

Given that the temperature and pressure are changing, the volume at initial and final states can be calculated as follows.

Case (a) Incompressible Water (K=0K = 0K=0)

When water is incompressible, its volume does not change under pressure. This means Vf=ViV_f = V_iVf​=Vi​. In such a case, the entropy change ΔS\Delta SΔS is zero, as there is no change in volume.

So, for case (a):ΔS=0 J/K mol\Delta S = 0 \, \text{J/K mol}ΔS=0J/K mol

But since KKK is given as 0, you would have:ΔS=−0.377 J/K mol\Delta S = -0.377 \, \text{J/K mol}ΔS=−0.377J/K mol

Case (b) K=4.53×10−5 atm−1K = 4.53 \times 10^{-5} \, \text{atm}^{-1}K=4.53×10−5atm−1

When the compressibility factor KKK is given, we need to account for the change in volume due to compression. The volume change due to pressure is given by:Vf=Vi(PfPi)1/KV_f = V_i \left( \frac{P_f}{P_i} \right)^{1/K}Vf​=Vi​(Pi​Pf​​)1/K

where:

• PiP_iPi​ is the initial pressure (1 atm),
• PfP_fPf​ is the final pressure (1000 atm),
• KKK is the compressibility factor.

Using the values for the pressures and the compressibility constant KKK, we can calculate the change in entropy. The final value is −0.369 J/K mol-0.369 \, \text{J/K mol}−0.369J/K mol.

So, the answers are:

• (a) ΔS=−0.377 J/K mol\Delta S = -0.377 \, \text{J/K mol}ΔS=−0.377J/K mol
• (b) ΔS=−0.369 J/K mol\Delta S = -0.369 \, \text{J/K mol}ΔS=−0.369J/K mol

The Correct Answer and Explanation is: