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 step by step, we will follow the principles of thermodynamics and the kinetic theory of gases.

### Step 1: Understanding the Given Information
We know that:
- The gas is helium, which is a monatomic ideal gas.
- The heat supplied (ΔQ) is 15 kJ.
- The process occurs at constant pressure.

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

### Step 3: Finding the Change in Internal Energy (ΔU)
For an ideal gas, the change in internal energy can be expressed in terms of heat added and the specific heat capacities. At constant pressure, we can relate the change in internal energy to the heat added using the ratio of specific heats (γ):
\[ \Delta U = \Delta Q \cdot \frac{C_v}{C_p} \]
Where:
- \( C_v \) is the specific heat at constant volume
- \( C_p \) is the specific heat at constant pressure

For a monatomic ideal gas like helium:
\[ \gamma = \frac{C_p}{C_v} = \frac{5}{3} \]
Thus:
\[ C_v = \frac{3}{2} R \quad \text{and} \quad C_p = \frac{5}{2} R \]

### Step 4: Calculating ΔU
Using the relationship:
\[ \Delta U = \Delta Q \cdot \frac{C_v}{C_p} = \Delta Q \cdot \frac{3/2 R}{5/2 R} = \Delta Q \cdot \frac{3}{5} \]

Substituting the value of ΔQ:
\[ \Delta U = 15 \text{ kJ} \cdot \frac{3}{5} = 9 \text{ kJ} \]

### Step 5: Finding the Work Done (ΔW)
Now, we can find the work done using the first law of thermodynamics:
\[ \Delta W = \Delta Q - \Delta U \]
Substituting the known values:
\[ \Delta W = 15 \text{ kJ} - 9 \text{ kJ} = 6 \text{ kJ} \]

### Final Results
- The variation of the internal energy of the gas (ΔU) is **9 kJ**.
- The work performed in the expansion (ΔW) is **6 kJ**.