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%`

Answer

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. **Understand the Process**:
- We have a monatomic gas expanding at constant pressure (isobaric process).
- The heat supplied (Q) is used for two purposes: increasing the internal energy (ΔU) and doing work (W) during expansion.

2. **Work Done in Isobaric Process**:
- The work done by the gas during expansion at constant pressure is given by:
\[ W = P \Delta V \]
- Using the ideal gas law, we can express this in terms of temperature:
\[ W = nR \Delta T \]
- Here, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( \Delta T \) is the change in temperature.

3. **Change in Internal Energy**:
- For an ideal gas, the change in internal energy is given by:
\[ \Delta U = n C_v \Delta T \]
- For a monatomic gas, the specific heat at constant volume \( C_v \) is:
\[ C_v = \frac{3}{2} R \]

4. **Heat Supplied in Isobaric Process**:
- The heat supplied at constant pressure is given by:
\[ Q = n C_p \Delta T \]
- For a monatomic gas, the specific heat at constant pressure \( C_p \) is:
\[ C_p = \frac{5}{2} R \]

5. **Calculate the Ratios**:
- Now, we need to find the ratios \( \frac{\Delta U}{Q} \) and \( \frac{W}{Q} \).
- First, calculate \( \frac{\Delta U}{Q} \):
\[ \frac{\Delta U}{Q} = \frac{n C_v \Delta T}{n C_p \Delta T} = \frac{C_v}{C_p} \]
- Since \( C_v = \frac{3}{2} R \) and \( C_p = \frac{5}{2} R \):
\[ \frac{\Delta U}{Q} = \frac{\frac{3}{2} R}{\frac{5}{2} R} = \frac{3}{5} \]
- This means that 60% of the heat supplied increases the internal energy of the gas.

6. **Calculate \( \frac{W}{Q} \)**:
- Now, calculate \( \frac{W}{Q} \):
\[ \frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p} \]
- Substituting \( C_p = \frac{5}{2} R \):
\[ \frac{W}{Q} = \frac{R}{\frac{5}{2} R} = \frac{2}{5} \]
- This means that 40% of the heat supplied is involved in the expansion.

### Final Results:
- Percentage of heat supplied that increases the internal energy: **60%**
- Percentage of heat supplied that is involved in the expansion: **40%**