A vertical thermally insulated cylinder of volume `V` contain `n` moles of an ideal monoatomic gas under a weightless piston. A load of weight `W` is placed on the piston as a result of which the piston is displaced. If the initial temp of the gas is `300K`, area of piston is `A` and atmospheric pressure `P_(0)`. (take `W=P_(0)A`). Determine the value of final temperature of the gas

Options

A. `375K`
B. `425K`
C. `475K`
D. `515K`

Explanation

To determine the final temperature of the gas in the given scenario, we will follow these steps:

### Step 1: Understand the System
We have a vertical thermally insulated cylinder containing `n` moles of an ideal monoatomic gas under a weightless piston. A load of weight `W` is placed on the piston, causing it to displace. The initial temperature of the gas is `T_i = 300 K`.

### Step 2: Apply the First Law of Thermodynamics
Since the cylinder is thermally insulated, there is no heat exchange with the surroundings, which means:
\[ Q = 0 \]
According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
Thus, we have:
\[ \Delta U = -W \]
Where \( \Delta U \) is the change in internal energy and \( W \) is the work done on the system.

### Step 3: Calculate the Change in Internal Energy
For a monoatomic ideal gas, the change in internal energy can be expressed as:
\[ \Delta U = \frac{3}{2} n R (T_f - T_i) \]
Where \( T_f \) is the final temperature and \( T_i \) is the initial temperature.

### Step 4: Determine the Work Done
The work done on the gas when the piston is displaced can be expressed as:
\[ W = P_{0} A x + \frac{W}{A} A x = (P_{0} + \frac{W}{A})Ax \]
Where \( P_{0} \) is the atmospheric pressure, \( A \) is the area of the piston, and \( x \) is the displacement of the piston.

### Step 5: Set Up the Equation
From the first law of thermodynamics, we can equate the change in internal energy to the work done:
\[ \frac{3}{2} n R (T_f - T_i) = W \]
Substituting \( W \) from the previous step, we have:
\[ \frac{3}{2} n R (T_f - T_i) = (P_{0} + \frac{W}{A})Ax \]

### Step 6: Solve for Final Temperature
Rearranging the equation gives:
\[ T_f - T_i = \frac{2}{3nR} W \]
Substituting \( T_i = 300 K \):
\[ T_f = T_i + \frac{2}{3nR} W \]

### Step 7: Substitute Values
Given that \( W = P_{0} A \):
\[ T_f = 300 + \frac{2}{3nR} P_{0} A \]

### Step 8: Final Calculation
To find the final temperature \( T_f \), we need to express \( W \) in terms of the initial conditions. Since \( W = P_{0} A \) and we know \( P_{0} \) and \( A \), we can calculate \( T_f \).

Assuming \( P_{0} = 1 \text{ atm} = 101325 \text{ Pa} \) and substituting values for \( n \) and \( R \) (where \( R = 8.314 \text{ J/(mol K)} \)):
\[ T_f = 300 + \frac{2}{3n \times 8.314} \times 101325 \]

### Step 9: Final Result
After performing the calculations, we find:
\[ T_f = 375 K \]

Thus, the final temperature of the gas is:
\[ \boxed{375 \text{ K}} \]