One mole of an ideal gas (`C_(V) = 20 JK^(-1) mol^(-1))` initially at STP is heated at constant volume to twice the initial temeprature. For the process W and q will be

Options

A. W = 0, q 5.46 kJ
B. W = 0 , q 0
C. W = -5.46kJ , q = 5.46 kJ
D. W = 5.46 kJ , q = 5.46 kJ

Answer

To solve the problem, we need to determine the work done (W) and the heat (q) for the process where one mole of an ideal gas is heated at constant volume from an initial temperature (T1) to twice that temperature (T2).

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Given:
- Moles of gas (n) = 1 mole
- Heat capacity at constant volume (C_v) = 20 J/K·mol
- Initial temperature (T1) = 273 K (Standard Temperature)
- Final temperature (T2) = 2 * T1 = 2 * 273 K = 546 K

2. **Calculate Change in Temperature (ΔT)**:
- ΔT = T2 - T1
- ΔT = 546 K - 273 K = 273 K

3. **Calculate Work Done (W)**:
- Since the process occurs at constant volume, the change in volume (ΔV) is 0.
- Work done (W) is given by the formula:
\[ W = P \Delta V \]
- Since ΔV = 0, we have:
\[ W = 0 \]

4. **Calculate Heat (q)**:
- The heat added (q) during the process at constant volume is given by:
\[ q = n C_v \Delta T \]
- Substituting the known values:
\[ q = 1 \, \text{mol} \times 20 \, \text{J/K·mol} \times 273 \, \text{K} \]
- Calculate:
\[ q = 20 \times 273 = 5460 \, \text{J} \]

5. **Convert Heat to Kilojoules**:
- To convert joules to kilojoules, divide by 1000:
\[ q = \frac{5460}{1000} = 5.46 \, \text{kJ} \]

### Final Results:
- Work done (W) = 0 J
- Heat added (q) = 5.46 kJ