Calulate the work done (in cal.). When 1.0 mole of `N_(2)H_(4)` decomposes completely against a pressure of `1.0` atm at `27^(@)C` `(Given R=2 cal//mol//K)`
`N_(2)H_(4)(l)toNH_(3)(g)+N_(2)(g)`

Options

A. `-1000cal`
B. `1000cal`
C. `-3000cal`
D. `3000cal`

Answer

To calculate the work done when 1 mole of \( N_2H_4 \) decomposes completely against a pressure of 1 atm at 27°C, we can follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition reaction of hydrazine (\( N_2H_4 \)) is:
\[ N_2H_4(l) \rightarrow NH_3(g) + N_2(g) \]
This indicates that 1 mole of \( N_2H_4 \) produces 1 mole of \( NH_3 \) and 1 mole of \( N_2 \).

### Step 2: Determine the change in the number of moles of gas (\( \Delta N_G \))
The change in the number of moles of gas is calculated as:
\[ \Delta N_G = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
From the balanced equation:
- Moles of gaseous products = 1 (from \( NH_3 \)) + 1 (from \( N_2 \)) = 2 moles
- Moles of gaseous reactants = 0 (since \( N_2H_4 \) is a liquid)

Thus,
\[ \Delta N_G = 2 - 0 = 2 \]

### Step 3: Use the formula for work done
The work done (\( W \)) is given by the formula:
\[ W = -\Delta N_G \cdot R \cdot T \]
Where:
- \( R = 2 \, \text{cal/mol/K} \)
- \( T = 27^\circ C = 27 + 273 = 300 \, \text{K} \)

### Step 4: Substitute the values into the work formula
Now substituting the values into the formula:
\[ W = -\Delta N_G \cdot R \cdot T = -2 \cdot 2 \, \text{cal/mol/K} \cdot 300 \, \text{K} \]
Calculating this gives:
\[ W = -2 \cdot 2 \cdot 300 = -1200 \, \text{cal} \]

### Final Answer
The work done when 1 mole of \( N_2H_4 \) decomposes completely against a pressure of 1 atm at 27°C is:
\[ W = -1200 \, \text{cal} \]

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