A monatomic gas expands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involed in the expansion is

Options

A. `75%, 25%`
B. `25%, 75%`
C. `60%, 40%`
D. `40%, 60%`

Explanation

To solve the problem, we need to determine the percentage of heat supplied that increases the internal energy of a monatomic gas and the percentage that is involved in the expansion when the gas expands at constant pressure.

### Step-by-Step Solution:

1. **Understanding the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ \Delta Q = \Delta U + \Delta W \]
where \(\Delta Q\) is the heat supplied, \(\Delta U\) is the change in internal energy, and \(\Delta W\) is the work done by the system.

2. **Expressing Heat Supplied at Constant Pressure**:
For a gas expanding at constant pressure, the heat supplied can be expressed as:
\[ \Delta Q = nC_p \Delta T \]
where \(n\) is the number of moles, \(C_p\) is the molar heat capacity at constant pressure, and \(\Delta T\) is the change in temperature.

3. **Change in Internal Energy for a Monatomic Gas**:
The change in internal energy for a monatomic gas is given by:
\[ \Delta U = nC_v \Delta T \]
where \(C_v\) is the molar heat capacity at constant volume.

4. **Relation Between \(C_p\) and \(C_v\)**:
For a monatomic gas, the relationship between \(C_p\) and \(C_v\) is:
\[ C_p = C_v + R \]
where \(R\) is the universal gas constant. For monatomic gases, the ratio \(\frac{C_p}{C_v} = \gamma\) and \(\gamma = \frac{5}{3}\). Thus, we can express \(C_v\) in terms of \(C_p\):
\[ C_v = \frac{3}{5}C_p \]

5. **Substituting \(C_v\) in the Internal Energy Equation**:
Now substituting \(C_v\) in the internal energy equation:
\[ \Delta U = n \left(\frac{3}{5}C_p\right) \Delta T \]

6. **Calculating Work Done**:
The work done during the expansion at constant pressure is:
\[ \Delta W = P \Delta V = nR \Delta T \]
Using the ideal gas law, we can express \(R\) in terms of \(C_p\):
\[ R = C_p - C_v = C_p - \frac{3}{5}C_p = \frac{2}{5}C_p \]
Therefore, the work done can also be expressed as:
\[ \Delta W = n \left(\frac{2}{5}C_p\right) \Delta T \]

7. **Substituting Back into the First Law**:
Now substituting \(\Delta U\) and \(\Delta W\) back into the first law:
\[ nC_p \Delta T = n\left(\frac{3}{5}C_p\right) \Delta T + n\left(\frac{2}{5}C_p\right) \Delta T \]
This confirms that:
\[ \Delta Q = \Delta U + \Delta W \]

8. **Calculating Percentages**:
- The percentage of heat supplied that increases the internal energy:
\[ \text{Percentage of } \Delta U = \frac{\Delta U}{\Delta Q} \times 100 = \frac{n\left(\frac{3}{5}C_p\right) \Delta T}{nC_p \Delta T} \times 100 = \frac{3}{5} \times 100 = 60\% \]
- The percentage of heat supplied that is involved in the expansion:
\[ \text{Percentage of } \Delta W = \frac{\Delta W}{\Delta Q} \times 100 = \frac{n\left(\frac{2}{5}C_p\right) \Delta T}{nC_p \Delta T} \times 100 = \frac{2}{5} \times 100 = 40\% \]

### Final Answer:
- The percentage of heat supplied that increases the internal energy of the gas is **60%**.
- The percentage of heat supplied that is involved in the expansion is **40%**.