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 step by step, we will analyze the situation of a monatomic gas expanding at constant pressure and determine the percentage of heat supplied that increases the internal energy of the gas and the percentage involved in the expansion.

### Step 1: Understand the process
The gas is expanding at constant pressure, which means it is undergoing an isobaric process. In this process, the heat supplied (ΔQ) 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.

### Step 2: Calculate the change in internal energy
The change in internal energy (ΔU) for a monatomic gas is given by:
\[ \Delta U = nC_v \Delta T \]
where \( C_v \) is the molar heat capacity at constant volume.

### Step 3: Apply the first law of thermodynamics
According to the first law of thermodynamics:
\[ \Delta Q = \Delta U + \Delta W \]
where \( \Delta W \) is the work done by the gas during expansion.

### Step 4: Express work done in terms of heat and internal energy
Rearranging the first law gives:
\[ \Delta W = \Delta Q - \Delta U \]

### Step 5: Calculate the percentage of heat that increases internal energy
We need to find the percentage of heat supplied that increases the internal energy:
\[ \text{Percentage of } \Delta U = \left( \frac{\Delta U}{\Delta Q} \right) \times 100\% \]
Substituting the expressions for ΔU and ΔQ:
\[ \text{Percentage of } \Delta U = \left( \frac{nC_v \Delta T}{nC_p \Delta T} \right) \times 100\% = \left( \frac{C_v}{C_p} \right) \times 100\% \]

### Step 6: Use values for a monatomic gas
For a monatomic gas:
- \( C_v = \frac{3R}{2} \)
- \( C_p = \frac{5R}{2} \)

Substituting these values:
\[ \frac{C_v}{C_p} = \frac{\frac{3R}{2}}{\frac{5R}{2}} = \frac{3}{5} \]
Thus:
\[ \text{Percentage of } \Delta U = \left( \frac{3}{5} \right) \times 100\% = 60\% \]

### Step 7: Calculate the percentage of heat involved in expansion
Now, we need to find the percentage of heat involved in the expansion:
\[ \text{Percentage of } \Delta W = \left( \frac{\Delta W}{\Delta Q} \right) \times 100\% \]
Substituting for ΔW:
\[ \Delta W = \Delta Q - \Delta U \]
So,
\[ \text{Percentage of } \Delta W = \left( \frac{\Delta Q - \Delta U}{\Delta Q} \right) \times 100\% = \left( 1 - \frac{\Delta U}{\Delta Q} \right) \times 100\% \]
Substituting the previously calculated percentage of ΔU:
\[ \text{Percentage of } \Delta W = \left( 1 - \frac{3}{5} \right) \times 100\% = \left( \frac{2}{5} \right) \times 100\% = 40\% \]

### Final Answer
Thus, the percentage of heat supplied that increases the internal energy of the gas is **60%**, and the percentage involved in the expansion is **40%**.