If heat is added at constant volume, 6300 J of heat are required to raise the temperature of an ideal gas by 150 K. If instead, heat is added at constant pressure, 8800 joules are required for the same temperature change. When the temperature of the gas changes by 300K, the internal energy of the gas changes by

Options

A. 5000 J
B. 12600 J
C. 17600 J
D. 22600 J

Answer

To solve the problem, we need to find the change in internal energy (\( \Delta U \)) of an ideal gas when the temperature changes by 300 K. We are given two scenarios: one at constant volume and one at constant pressure.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- At constant volume, the heat added (\( Q_V \)) to raise the temperature by 150 K is 6300 J.
- At constant pressure, the heat added (\( Q_P \)) to raise the temperature by 150 K is 8800 J.
- We need to find the change in internal energy when the temperature changes by 300 K.

2. **Using the First Law of Thermodynamics**:
The first law states:
\[ \Delta Q = \Delta U + \Delta W \]
- At constant volume, \( \Delta W = 0 \), so:
\[ Q_V = \Delta U \]
- Therefore, for the first scenario:
\[ \Delta U = Q_V = 6300 \text{ J} \]

3. **Finding the Change in Internal Energy for 300 K**:
- The change in internal energy is proportional to the change in temperature. Since we have the change in internal energy for a 150 K change, we can find it for a 300 K change.
- The relationship can be expressed as:
\[ \frac{\Delta U_2}{\Delta U_1} = \frac{\Delta T_2}{\Delta T_1} \]
where:
- \( \Delta U_1 = 6300 \text{ J} \) (for \( \Delta T_1 = 150 \text{ K} \))
- \( \Delta T_2 = 300 \text{ K} \)

4. **Calculating \( \Delta U_2 \)**:
- Plugging in the values:
\[ \frac{\Delta U_2}{6300} = \frac{300}{150} \]
- Simplifying:
\[ \frac{\Delta U_2}{6300} = 2 \]
- Therefore:
\[ \Delta U_2 = 6300 \times 2 = 12600 \text{ J} \]

5. **Final Answer**:
The change in internal energy of the gas when the temperature changes by 300 K is:
\[ \Delta U_2 = 12600 \text{ J} \]