A contribution of both heat (enthalpy) and randomness(entropy) shall be considered to the overall spontaneity of process, when deciding the spontaneity of a chemical reaction or other process, we define a quantity called the Gibb's energy change `(DeltaG)`.
`DeltaG=DeltaH-TDeltaS`
where,`DeltaH`=Enthalpy change, `DeltaS`=Entropy change , T=Temperature in kelvin.
If `DeltaGlt0`, Process is spontaneous , `DeltaG=0`, Process is at equilibrium , `DeltaGgt0`, Process is non-spontaneous.
5 mol of liquid water is compressed from 1 bar to 10 bar at constant temperature.change in Gibb's energy `(DeltaG)` in Joule is:[Density of water =`1000 kg//m^3`]

Options

A. 18
B. 225
C. 450
D. 900

Explanation

To solve the problem of calculating the change in Gibbs energy (ΔG) when 5 moles of liquid water is compressed from 1 bar to 10 bar at constant temperature, we will follow these steps:

### Step 1: Understand the Given Information
- Number of moles of water (n) = 5 mol
- Initial pressure (P1) = 1 bar
- Final pressure (P2) = 10 bar
- Density of water (ρ) = 1000 kg/m³
- Temperature change (ΔT) = 0 (constant temperature)

### Step 2: Calculate the Mass of Water
To find the mass of water, we use the molar mass of water:
- Molar mass of water (H2O) = 18 g/mol

Now, calculate the mass (m):
\[ m = n \times \text{Molar mass} = 5 \, \text{mol} \times 18 \, \text{g/mol} = 90 \, \text{g} \]
Convert grams to kilograms:
\[ m = 90 \, \text{g} = 0.090 \, \text{kg} \]

### Step 3: Calculate the Volume of Water
Using the formula for volume (V):
\[ V = \frac{m}{\rho} \]
Substituting the values:
\[ V = \frac{0.090 \, \text{kg}}{1000 \, \text{kg/m}^3} = 9.0 \times 10^{-5} \, \text{m}^3 \]

### Step 4: Calculate the Change in Pressure (ΔP)
\[ \Delta P = P2 - P1 = 10 \, \text{bar} - 1 \, \text{bar} = 9 \, \text{bar} \]
Convert bar to Pascal (1 bar = 100,000 Pa):
\[ \Delta P = 9 \, \text{bar} \times 100,000 \, \text{Pa/bar} = 900,000 \, \text{Pa} \]

### Step 5: Calculate the Change in Gibbs Energy (ΔG)
Using the equation:
\[ \Delta G = V \Delta P \]
Substituting the values:
\[ \Delta G = (9.0 \times 10^{-5} \, \text{m}^3) \times (900,000 \, \text{Pa}) = 81 \, \text{J} \]

### Step 6: Final Calculation
Since we need to calculate ΔG in Joules:
\[ \Delta G = 81 \, \text{J} \]

### Conclusion
The change in Gibbs energy (ΔG) when 5 moles of liquid water is compressed from 1 bar to 10 bar at constant temperature is **81 Joules**.