iIn isochoric process change ib internal energy of the gas in `triangle` T temperature rise for 2 mole is

Options

A. `2C_p/_\T`
B. `C_V/2/_\T`
C. `2C_V/_\T`
D. Zero

Answer

To find the change in internal energy of a gas during an isochoric process for 2 moles with a temperature rise of ΔT, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Isochoric Process**:
- An isochoric process is one in which the volume of the gas remains constant. Therefore, no work is done on or by the gas.

2. **Formula for Change in Internal Energy**:
- The change in internal energy (ΔU) for an ideal gas during an isochoric process can be expressed as:
\[ \Delta U = n C_V \Delta T \]
where:
- \( n \) = number of moles of gas
- \( C_V \) = specific heat at constant volume
- \( \Delta T \) = change in temperature

3. **Substituting Given Values**:
- From the problem, we know:
- \( n = 2 \) moles
- \( \Delta T = 2 \) (the temperature rise)
- We can substitute these values into the formula:
\[ \Delta U = 2 C_V \cdot 2 \]

4. **Simplifying the Expression**:
- Now, simplify the expression:
\[ \Delta U = 4 C_V \]

5. **Final Result**:
- The change in internal energy for the gas during the isochoric process is:
\[ \Delta U = 4 C_V \]