A certain amount of an ideal monatomic gas needs 20 J of heat energy to raise its temperature by 10°C at constant pressure. The heat needed for the same temperature rise at constant volume will be

Options

A. 30 j
B. 12 j
C. 200 j
D. 2153 j

Answer

To solve the problem, we need to find the heat required to raise the temperature of an ideal monatomic gas by 10°C at constant volume, given that it requires 20 J of heat at constant pressure for the same temperature change.

### Step-by-step Solution:

1. **Understand the Given Information**:
- Heat required at constant pressure (Q_p) = 20 J
- Temperature change (ΔT) = 10°C
- The gas is monatomic.

2. **Use the Formula for Heat at Constant Pressure**:
The heat absorbed by the gas at constant pressure is given by:
\[ Q_p = N C_p \Delta T \]
where:
- \(N\) = number of moles of the gas
- \(C_p\) = molar specific heat at constant pressure
- \(\Delta T\) = change in temperature

3. **Substituting Known Values**:
We know that \(Q_p = 20 J\) and \(\Delta T = 10°C\). For a monatomic ideal gas, the molar specific heat at constant pressure is:
\[ C_p = \frac{5R}{2} \]
Therefore, we can write:
\[ 20 = N \left(\frac{5R}{2}\right) (10) \]

4. **Simplifying the Equation**:
Rearranging the equation gives:
\[ 20 = 25NR \implies NR = \frac{20}{25} = \frac{4}{5} \]

5. **Use the Formula for Heat at Constant Volume**:
The heat absorbed by the gas at constant volume is given by:
\[ Q_v = N C_v \Delta T \]
where for a monatomic ideal gas, the molar specific heat at constant volume is:
\[ C_v = \frac{3R}{2} \]

6. **Substituting into the Heat at Constant Volume Formula**:
Substituting \(C_v\) and \(\Delta T\):
\[ Q_v = N \left(\frac{3R}{2}\right) (10) \]

7. **Substituting \(NR\) Value**:
We already found \(NR = \frac{4}{5}\):
\[ Q_v = \left(\frac{4}{5}\right) \left(\frac{3R}{2}\right) (10) \]
Simplifying gives:
\[ Q_v = \frac{4 \cdot 3R \cdot 10}{5 \cdot 2} = \frac{120R}{10} = 12R \]

8. **Final Calculation**:
Since \(R\) is a constant, we can conclude that the heat required at constant volume for the same temperature change is:
\[ Q_v = 12 J \]

### Final Answer:
The heat needed for the same temperature rise at constant volume is **12 J**.