A vertical hollow cylinder contains an ideal gas. The gas is enclosed by a `5kg` movable piston with an area of cross-section `5 xx 10^(-3)m^(2)`. Now, the gas is heated slowly from `300K` to `350K` and the piston rises by `0.1m`. The piston is now clamped at this position and the gas is cooled back to `300K`. Fid the difference between the heat energy added during heating process and energy lost during the cooling process. `("1atm pressire" = 10^(5)Nm^(-2)]`

Explanation

To solve the problem, we need to find the difference between the heat energy added during the heating process and the energy lost during the cooling process of the ideal gas in the vertical hollow cylinder with a movable piston.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the piston, \( m = 5 \, \text{kg} \)
- Area of cross-section of the piston, \( A = 5 \times 10^{-3} \, \text{m}^2 \)
- Initial temperature, \( T_1 = 300 \, \text{K} \)
- Final temperature, \( T_2 = 350 \, \text{K} \)
- Displacement of the piston, \( h = 0.1 \, \text{m} \)
- Atmospheric pressure, \( P_0 = 10^5 \, \text{N/m}^2 \)

2. **Calculate the Work Done by the Gas:**
The work done \( W \) by the gas during the heating process when the piston rises can be calculated using the formula:
\[ W = P \Delta V \]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.

The change in volume \( \Delta V \) can be calculated as:
\[ \Delta V = A \cdot h = (5 \times 10^{-3} \, \text{m}^2) \cdot (0.1 \, \text{m}) = 5 \times 10^{-4} \, \text{m}^3 \]

3. **Calculate the Pressure Acting on the Piston:**
The pressure \( P \) acting on the piston is the sum of the atmospheric pressure and the pressure due to the weight of the piston:
\[ P = P_0 + \frac{mg}{A} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
\[ P = 10^5 + \frac{(5)(9.81)}{5 \times 10^{-3}} = 10^5 + 9810 = 109810 \, \text{N/m}^2 \]

4. **Calculate the Work Done:**
Now substituting the values into the work done formula:
\[ W = P \Delta V = (109810 \, \text{N/m}^2)(5 \times 10^{-4} \, \text{m}^3) = 54.905 \, \text{J} \]

5. **Determine Heat Energy Added:**
Since the process is quasi-static and the internal energy change is zero, the heat added \( Q \) during heating is equal to the work done:
\[ Q_{\text{added}} = W = 54.905 \, \text{J} \approx 55 \, \text{J} \]

6. **Cooling Process:**
When the gas is cooled back to \( 300 \, \text{K} \), since the piston is clamped, the volume remains constant. Therefore, no work is done during the cooling process, and the energy lost \( Q_{\text{lost}} \) is equal to the heat removed from the gas, which is also equal to the internal energy change.

7. **Calculate the Difference:**
Since the work done during heating is equal to the heat added and the work done during cooling is zero:
\[ \text{Difference} = Q_{\text{added}} - Q_{\text{lost}} = 55 \, \text{J} - 0 \, \text{J} = 55 \, \text{J} \]

### Final Answer:
The difference between the heat energy added during the heating process and the energy lost during the cooling process is **55 J**.