1 mol of `NH_(3)` gas`( gamma = 1.33)` at `27^(@)C`is allowed to expand adiabatically so that the final volume becomes 8 times. Work done will be

Options

A. 450 cal
B. 1800 cal
C. 900 cal
D. 300 cal

Explanation

To solve the problem of work done during the adiabatic expansion of 1 mole of `NH3` gas, we will follow these steps:

### Step 1: Convert the initial temperature to Kelvin
The initial temperature \( T_1 \) is given as \( 27^{\circ}C \).
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
Thus,
\[ T_1 = 27 + 273.15 = 300.15 \approx 300 \, K \]

### Step 2: Identify the final volume
The problem states that the final volume \( V_2 \) is 8 times the initial volume \( V_1 \). Therefore,
\[ V_2 = 8V_1 \]

### Step 3: Use the adiabatic relation to find the final temperature
For an adiabatic process, the relationship between temperature and volume is given by:
\[ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]
Substituting \( V_2 = 8V_1 \) and \( \gamma = 1.33 \):
\[ \frac{T_2}{300} = \left(\frac{V_1}{8V_1}\right)^{1.33 - 1} = \left(\frac{1}{8}\right)^{0.33} \]
Calculating \( \left(\frac{1}{8}\right)^{0.33} \):
\[ \left(\frac{1}{8}\right)^{0.33} = \left(8^{-1}\right)^{0.33} = 8^{-0.33} = 2^{-1} = \frac{1}{2} \]
Thus,
\[ \frac{T_2}{300} = \frac{1}{2} \implies T_2 = 150 \, K \]

### Step 4: Calculate the change in internal energy
The change in internal energy \( \Delta U \) for an ideal gas is given by:
\[ \Delta U = n C_v \Delta T \]
Where:
- \( n = 1 \, \text{mol} \)
- \( C_v = \frac{R}{\gamma - 1} \)
- \( \Delta T = T_2 - T_1 = 150 - 300 = -150 \, K \)

### Step 5: Calculate \( C_v \)
Using the value of \( R \) (approximately \( 2 \, \text{cal/mol K} \)):
\[ C_v = \frac{R}{\gamma - 1} = \frac{2}{1.33 - 1} = \frac{2}{0.33} \approx 6.06 \, \text{cal/mol K} \]

### Step 6: Calculate \( \Delta U \)
Now substituting the values into the equation for \( \Delta U \):
\[ \Delta U = 1 \times 6.06 \times (-150) = -909 \, \text{cal} \]

### Step 7: Relate \( \Delta U \) to work done
In an adiabatic process, the work done \( W \) is equal to the change in internal energy:
\[ W = \Delta U = -909 \, \text{cal} \]

### Final Answer
The work done during the adiabatic expansion is:
\[ W = -909 \, \text{cal} \]