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.

Options

A. (a)zero
B. (b)`136 J`
C. (c)`- 136 J`
D. (d)`- 68 J`

Explanation

To solve the problem, we need to analyze the heating and cooling process of the ideal gas in the cylinder with a movable piston. We will calculate the heat added to the gas during heating (ΔQ1) and the heat lost during cooling (ΔQ2), and then find the difference between these two quantities.

### Step-by-Step Solution:

1. **Calculate the Total Pressure on the Gas:**
The total pressure (P) on the gas is the sum of the atmospheric pressure and the pressure due to the weight of the piston.
\[ P = P_{\text{atm}} + \frac{mg}{A} \]
Where:
- \( P_{\text{atm}} = 100 \, \text{kPa} = 100 \times 10^3 \, \text{Pa} \)
- \( m = 8 \, \text{kg} \) (mass of the piston)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( A = 60 \, \text{cm}^2 = 60 \times 10^{-4} \, \text{m}^2 \) (area of the piston)

Calculate the pressure due to the weight of the piston:
\[ P_{\text{piston}} = \frac{mg}{A} = \frac{8 \times 9.8}{60 \times 10^{-4}} = \frac{78.4}{0.006} = 13067 \, \text{Pa} = 13.067 \, \text{kPa} \]

Now, calculate the total pressure:
\[ P = 100 \times 10^3 + 13.067 \times 10^3 = 113067 \, \text{Pa} = 113.067 \, \text{kPa} \]

2. **Calculate the Change in Volume (ΔV):**
The piston rises by 20 cm, which is:
\[ \Delta V = A \times h = 60 \times 10^{-4} \, \text{m}^2 \times 0.2 \, \text{m} = 12 \times 10^{-4} \, \text{m}^3 = 0.0012 \, \text{m}^3 \]

3. **Calculate the Work Done (W) during Heating:**
The work done by the gas when the piston moves is given by:
\[ W = P \Delta V = 113067 \, \text{Pa} \times 0.0012 \, \text{m}^3 = 135.68 \, \text{J} \]

4. **Calculate the Change in Internal Energy (ΔU) during Heating:**
The change in internal energy for an ideal gas can be calculated using:
\[ \Delta U = nC_v \Delta T \]
Where:
- \( C_v \) is the specific heat at constant volume.
- \( \Delta T = (100 - 30) \, \text{°C} = 70 \, \text{°C} = 70 \, \text{K} \)

We need the number of moles (n) of the gas, which can be calculated using the ideal gas law:
\[ PV = nRT \]
Rearranging gives:
\[ n = \frac{PV}{RT} \]
Assuming \( R = 8.314 \, \text{J/(mol K)} \) and using an average temperature of \( 65 \, \text{°C} = 338.15 \, \text{K} \):
\[ n = \frac{113067 \times 0.0012}{8.314 \times 338.15} = \frac{135.68}{2814.67} \approx 0.048 \, \text{mol} \]

Now, we can calculate ΔU:
\[ \Delta U = nC_v \Delta T \quad \text{(assuming } C_v \text{ for a diatomic gas is about } 5R/2) \]
\[ C_v \approx 5 \times 8.314 / 2 = 20.785 \, \text{J/(mol K)} \]
\[ \Delta U = 0.048 \times 20.785 \times 70 \approx 70.0 \, \text{J} \]

5. **Calculate the Heat Added (ΔQ1):**
Using the first law of thermodynamics:
\[ \Delta Q_1 = \Delta U + W = 70.0 \, \text{J} + 135.68 \, \text{J} \approx 205.68 \, \text{J} \]

6. **Calculate the Heat Lost (ΔQ2) during Cooling:**
When the gas is cooled back to 30°C, the process is at constant volume, so:
\[ \Delta Q_2 = \Delta U = nC_v \Delta T = 0.048 \times 20.785 \times (-70) \approx -70.0 \, \text{J} \]

7. **Find the Difference (ΔQ1 - ΔQ2):**
\[ \Delta Q = \Delta Q_1 - \Delta Q_2 = 205.68 - (-70.0) = 205.68 + 70.0 = 275.68 \, \text{J} \]

### Final Answer:
The difference in heat added and heat lost is approximately **275.68 J**.