Temperature of four moles of a monoatomic gas is increased by 300K in isochoric process. Find W, Q and `DeltaU`.

Explanation

To solve the problem, we need to find the work done (W), heat added (Q), and the change in internal energy (ΔU) for a monoatomic gas undergoing an isochoric process where the temperature increases by 300 K.

### Step-by-Step Solution:

1. **Identify the Process**:
- The process is isochoric, meaning the volume remains constant.

2. **Calculate Work Done (W)**:
- In an isochoric process, the work done is given by the formula:
\[ W = \int P \, dV \]
- Since the volume does not change (dV = 0), the work done is:
\[ W = 0 \]

3. **Use the First Law of Thermodynamics**:
- The first law of thermodynamics states:
\[ \Delta U = Q - W \]
- Since we have already established that \(W = 0\), we can simplify this to:
\[ \Delta U = Q \]

4. **Calculate Change in Internal Energy (ΔU)**:
- The change in internal energy for a monoatomic ideal gas can be calculated using the formula:
\[ \Delta U = n C_V \Delta T \]
- For a monoatomic gas, the specific heat at constant volume \(C_V\) is:
\[ C_V = \frac{3}{2} R \]
- Given:
- Number of moles, \(n = 4\)
- Change in temperature, \(\Delta T = 300 \, K\)
- Plugging in the values:
\[ \Delta U = 4 \times \left(\frac{3}{2} R\right) \times 300 \]
- Simplifying:
\[ \Delta U = 4 \times \frac{3}{2} \times 300 \times R = 1800 R \]

5. **Calculate Heat Added (Q)**:
- Since we established earlier that \(Q = \Delta U\) (because \(W = 0\)):
\[ Q = 1800 R \]

### Summary of Results:
- Work Done, \(W = 0\)
- Heat Added, \(Q = 1800 R\)
- Change in Internal Energy, \(\Delta U = 1800 R\)