`2CaSO_4 (s) hArr 2CaO(s)+2SO_(2)(g)+O_2(g), DeltaHgt0`
Above equilibrium is establised by taking some amount of `CaSO_4(s)` in a closed container at 1600 K. Then which of the following may be correct options ?

Options

A. Moles of CaO(s) will increase with the increase in temperature
B. if the volume of the container is doubled at equilibrium then partial pressure of `SO_2(g)` will change at new equilibrium
C. If the volume of the container is halved partial pressure of `O_2`(g) at new equilibrium will remain same
D. If two moles of the He gas is added at constant pressure then the moles of CaO(s) will increase

Explanation

To solve the question, we need to analyze the given equilibrium reaction:

\[ 2 \text{CaSO}_4 (s) \rightleftharpoons 2 \text{CaO}(s) + 2 \text{SO}_2(g) + \text{O}_2(g) \]

with \(\Delta H > 0\), indicating that the reaction is endothermic.

### Step 1: Understand the Effect of Temperature
Since the reaction is endothermic, increasing the temperature will favor the formation of products (CaO, SO2, and O2) according to Le Chatelier's principle.

**Conclusion**: As the temperature increases, the moles of calcium oxide (CaO) will increase.

### Step 2: Analyze the Effect of Volume Change on Partial Pressure
If the volume of the container is doubled, the equilibrium constant \(K_p\) is dependent only on temperature and not on the volume of the container. Therefore, the partial pressure of sulfur dioxide (SO2) will remain unchanged.

**Conclusion**: The partial pressure of SO2 will not change when the volume of the container is doubled.

### Step 3: Analyze the Effect of Halving the Volume
If the volume of the container is halved, the equilibrium will shift to the side with fewer moles of gas. In this case, the products side has 3 moles of gas (2 SO2 + 1 O2) and the reactants side has 0 moles of gas. Hence, the equilibrium will shift towards the products, and the partial pressure of oxygen (O2) will remain constant because it depends on temperature.

**Conclusion**: The partial pressure of O2 will remain the same when the volume is halved.

### Step 4: Effect of Adding Helium Gas
Adding helium gas at constant pressure will increase the total volume. The equilibrium will shift towards the side with more moles of gas, which is the products side in this case. Therefore, the moles of calcium oxide (CaO) will increase.

**Conclusion**: The moles of calcium oxide will increase if 2 moles of helium gas are added at constant pressure.

### Final Summary of Correct Options
1. Moles of calcium oxide will increase with an increase in temperature.
2. The partial pressure of sulfur dioxide will not change when the volume is doubled.
3. The partial pressure of oxygen will remain the same when the volume is halved.
4. Moles of calcium oxide will increase if helium gas is added at constant pressure.

Thus, options 1, 3, and 4 are correct.