One litre of dry air STP undergoes through an adiabatic process and reaches to a final temperature of 89.6 °C, the work done by air is : [Take air to be an ideal gas and `gamma_("air")=1.4` use R`=8.3J mol^(-1)K^(-1)]`

Options

A. 83J
B. -83J
C. 1859.6J
D. -1859.2J

Explanation

To solve the problem of work done by 1 litre of dry air undergoing an adiabatic process, we can follow these steps:

### Step 1: Understand the Initial Conditions
At Standard Temperature and Pressure (STP), the initial temperature \( T_i \) is 0 °C, which is equivalent to 273 K. The volume \( V \) is given as 1 litre, which can be converted to cubic meters:
\[ V = 1 \, \text{litre} = 1 \times 10^{-3} \, \text{m}^3 \]

### Step 2: Calculate the Final Temperature
The final temperature \( T_f \) is given as 89.6 °C. We need to convert this to Kelvin:
\[ T_f = 89.6 + 273 = 362.6 \, \text{K} \]

### Step 3: Calculate the Change in Temperature
The change in temperature \( \Delta T \) can be calculated as:
\[ \Delta T = T_f - T_i = 362.6 \, \text{K} - 273 \, \text{K} = 89.6 \, \text{K} \]

### Step 4: Calculate the Number of Moles of Air
Using the ideal gas law, we can find the number of moles \( n \):
\[ PV = nRT \]
Where:
- \( P = 10^5 \, \text{Pa} \) (pressure at STP)
- \( R = 8.3 \, \text{J/(mol K)} \)
- \( T = T_i = 273 \, \text{K} \)

Rearranging the equation gives:
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(10^5 \, \text{Pa})(1 \times 10^{-3} \, \text{m}^3)}{(8.3 \, \text{J/(mol K)})(273 \, \text{K})} \]
Calculating this gives:
\[ n \approx \frac{100}{2270.9} \approx 0.044 \, \text{mol} \]

### Step 5: Calculate \( nR \)
Now, we calculate \( nR \):
\[ nR = n \times R = 0.044 \, \text{mol} \times 8.3 \, \text{J/(mol K)} \approx 0.3652 \, \text{J/K} \]

### Step 6: Calculate the Work Done in the Adiabatic Process
The work done \( W \) in an adiabatic process can be calculated using the formula:
\[ W = \frac{nR \Delta T}{1 - \gamma} \]
Where \( \gamma = 1.4 \) for air. Thus:
\[ W = \frac{0.3652 \, \text{J/K} \times 89.6 \, \text{K}}{1 - 1.4} \]
Calculating the denominator:
\[ 1 - 1.4 = -0.4 \]
Now substituting:
\[ W = \frac{0.3652 \times 89.6}{-0.4} \]
Calculating the numerator:
\[ 0.3652 \times 89.6 \approx 32.7 \]
Thus:
\[ W \approx \frac{32.7}{-0.4} \approx -81.75 \, \text{J} \]

### Step 7: Conclusion
The work done by the air is approximately -81.75 J. The negative sign indicates that work is done by the system.