130 g of Zn is dissolved in dilute sulphuric acid in an open beaker . Find the work done in the process assuming isothermal operation at 300K.

Options

A. `-1200` cal
B. `-1800` cal
C. `+1800` cal
D. none

Explanation

To solve the problem of finding the work done when 130 g of zinc is dissolved in dilute sulfuric acid at 300 K, we can follow these steps:

### Step 1: Write the Reaction Equation
The reaction of zinc with dilute sulfuric acid can be represented as:
\[ \text{Zn (s)} + \text{H}_2\text{SO}_4 (aq) \rightarrow \text{ZnSO}_4 (aq) + \text{H}_2 (g) \]

### Step 2: Calculate the Number of Moles of Zinc
To find the number of moles of zinc (Zn), we use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]

Given:
- Mass of Zn = 130 g
- Molar mass of Zn = 65.38 g/mol

Calculating the number of moles:
\[ n = \frac{130 \, \text{g}}{65.38 \, \text{g/mol}} \approx 1.99 \, \text{moles} \]
This can be approximated to 2 moles.

### Step 3: Determine the Change in Moles of Gas (ΔNg)
In the reaction, 1 mole of zinc produces 1 mole of hydrogen gas (H₂). Therefore, the change in moles of gas (ΔNg) is:
\[ \Delta N_g = \text{moles of products} - \text{moles of reactants} \]
\[ \Delta N_g = 1 \, \text{(from H}_2\text{)} - 0 \, \text{(from Zn)} = 1 \]

### Step 4: Use the Formula for Work Done
The work done (W) in an isothermal process can be calculated using the formula:
\[ W = -\Delta N_g \cdot R \cdot T \]

Where:
- R = 2 cal/K·mol (universal gas constant)
- T = 300 K

Substituting the values:
\[ W = -1 \, \text{mol} \cdot 2 \, \text{cal/K·mol} \cdot 300 \, \text{K} \]
\[ W = -600 \, \text{cal} \]

### Step 5: Final Result
The work done in the process is:
\[ W = -600 \, \text{cal} \]