A cylinder of ideal gas is closed by an 8kg movable piston of area `60cm^2`. The atmospheric pressure is 100kPa. When the gas is heated form `30^@C` to `100^@C`, the piston rises 20 cm. The piston is then fastened in the place and the gas is cooled back to `30^@C`. If `DeltaQ_1` is the heat added to the gas during heating and `DeltaQ_2` is the heat lost during cooling, find the difference.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
- Mass of the piston, \( m = 8 \, \text{kg} \)
- Area of the piston, \( A = 60 \, \text{cm}^2 = 60 \times 10^{-4} \, \text{m}^2 \)
- Atmospheric pressure, \( P_{\text{atm}} = 100 \, \text{kPa} = 10^5 \, \text{Pa} \)
- Initial temperature, \( T_1 = 30^\circ C = 303 \, \text{K} \)
- Final temperature, \( T_2 = 100^\circ C = 373 \, \text{K} \)
- Displacement of the piston, \( x = 20 \, \text{cm} = 0.2 \, \text{m} \)

### Step 2: Calculate the Pressure of the Gas
The pressure of the gas can be calculated using the equation of forces acting on the piston:
\[ P_g = P_{\text{atm}} + \frac{mg}{A} \]
Where \( g \approx 9.81 \, \text{m/s}^2 \).

Substituting the values:
\[ P_g = 10^5 + \frac{8 \times 9.81}{60 \times 10^{-4}} \]
Calculating:
\[ P_g = 10^5 + \frac{78.48}{0.006} = 10^5 + 13080 = 113080 \, \text{Pa} \]

### Step 3: Calculate the Work Done During Heating
Since the process is at constant pressure, the work done \( W \) is given by:
\[ W = P_g \cdot A \cdot x \]
Substituting the values:
\[ W = 113080 \times 60 \times 10^{-4} \times 0.2 \]
Calculating:
\[ W = 113080 \times 0.006 \times 0.2 = 1356.96 \, \text{J} \]

### Step 4: Calculate the Heat Added \( \Delta Q_1 \)
Using the first law of thermodynamics:
\[ \Delta Q_1 = \Delta U + W \]
Where:
\[ \Delta U = n C_v \Delta T \]
Given \( \Delta T = T_2 - T_1 = 373 - 303 = 70 \, \text{K} \).

Thus:
\[ \Delta Q_1 = n C_v \cdot 70 + 1356.96 \]

### Step 5: Calculate the Heat Lost During Cooling \( \Delta Q_2 \)
When the piston is fixed, the work done is zero, so:
\[ \Delta Q_2 = \Delta U \]
Where:
\[ \Delta U = n C_v \cdot (-70) = -70 n C_v \]
Thus:
\[ \Delta Q_2 = 70 n C_v \]

### Step 6: Find the Difference \( \Delta Q_1 - \Delta Q_2 \)
Now, we can find the difference:
\[ \Delta Q_1 - \Delta Q_2 = (70 n C_v + 1356.96) - (70 n C_v) = 1356.96 \, \text{J} \]

### Final Answer
The difference \( \Delta Q_1 - \Delta Q_2 \) is approximately:
\[ \Delta Q_1 - \Delta Q_2 = 1356.96 \, \text{J} \]