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. `83 J`
B. `-83 J`
C. `1859.2 J`
D. `-1859.2 J`

Explanation

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

### Step 1: Identify the initial conditions
- The initial temperature (T_initial) is given as 0 °C, which is equivalent to 273 K.
- The final temperature (T_final) is given as 89.6 °C, which is equivalent to 89.6 + 273 = 362.6 K.
- The volume of air (V) is 1 liter = 1 × 10^(-3) m³.
- The pressure at STP (P_initial) is 10^5 Pa.

### Step 2: Calculate the number of moles (n) of air
Using the ideal gas law:
\[ PV = nRT \]
Where:
- P = 10^5 Pa
- V = 1 × 10^(-3) m³
- R = 8.3 J/(mol·K)
- T_initial = 273 K

Rearranging the equation to solve for n:
\[ n = \frac{PV}{RT} \]
Substituting the known values:
\[ n = \frac{(10^5)(1 \times 10^{-3})}{(8.3)(273)} \]

Calculating:
\[ n = \frac{100}{2267.9} \approx 0.0441 \text{ moles} \]

### Step 3: Use the adiabatic process formula for work done
For an adiabatic process, the work done (W) can be calculated using the formula:
\[ W = \frac{nR(T_{final} - T_{initial})}{1 - \gamma} \]
Where:
- \(\gamma\) (gamma) for air = 1.4

Substituting the values:
\[ W = \frac{0.0441 \times 8.3 \times (362.6 - 273)}{1 - 1.4} \]

Calculating the change in temperature:
\[ T_{final} - T_{initial} = 362.6 - 273 = 89.6 \text{ K} \]

Now substituting this into the work done equation:
\[ W = \frac{0.0441 \times 8.3 \times 89.6}{1 - 1.4} \]
\[ W = \frac{0.0441 \times 8.3 \times 89.6}{-0.4} \]

Calculating the numerator:
\[ 0.0441 \times 8.3 \times 89.6 \approx 33.3 \]

Now substituting this back into the work done equation:
\[ W = \frac{33.3}{-0.4} \approx -83.25 \text{ J} \]

### Step 4: Final answer
The work done by the air during the adiabatic process is approximately:
\[ W \approx -83.25 \text{ J} \]