A Vessel contains helium, which expands at constant pressure when 15 kJ of heat is supplied to it. What will be the variation of the internal energy of the gas? What is the work performed in the expansion?

Explanation

To solve the problem, we will use the principles of thermodynamics related to the internal energy, heat transfer, and work done during the expansion of a gas at constant pressure.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Heat supplied (ΔQ) = 15 kJ
- The gas is helium, which is a monatomic ideal gas.
- The process occurs at constant pressure.

2. **Using the First Law of Thermodynamics**:
The first law of thermodynamics states that:
\[ \Delta Q = \Delta U + \Delta W \]
Where:
- ΔQ = heat added to the system
- ΔU = change in internal energy
- ΔW = work done by the system

3. **Work Done in Constant Pressure Process**:
For an ideal gas expanding at constant pressure, the work done (ΔW) can be expressed as:
\[ \Delta W = P \Delta V \]
However, we will also relate it to the change in internal energy and heat supplied.

4. **Relating ΔQ, ΔU, and ΔW**:
We can express the relationships in terms of the specific heat capacities. For a monatomic gas like helium:
- \( C_p = \frac{5}{2} R \)
- \( C_v = \frac{3}{2} R \)
- The ratio \( \gamma = \frac{C_p}{C_v} = \frac{5/2}{3/2} = \frac{5}{3} \)

5. **Finding the Ratios**:
From the relationships, we can derive:
\[ \frac{\Delta Q}{\Delta U} = \frac{5}{3}, \quad \frac{\Delta Q}{\Delta W} = \frac{5}{2} \]
This implies:
\[ \Delta U = \frac{3}{5} \Delta Q \]
\[ \Delta W = \frac{2}{5} \Delta Q \]

6. **Calculating ΔU**:
Substituting ΔQ = 15 kJ into the equation for ΔU:
\[ \Delta U = \frac{3}{5} \times 15 \text{ kJ} = 9 \text{ kJ} \]

7. **Calculating ΔW**:
Substituting ΔQ = 15 kJ into the equation for ΔW:
\[ \Delta W = \frac{2}{5} \times 15 \text{ kJ} = 6 \text{ kJ} \]

### Final Answers:
- Variation of the internal energy of the gas (ΔU) = 9 kJ
- Work performed in the expansion (ΔW) = 6 kJ